Physics
98 Solved Past Papers
2008 – 2024 Archives
Nuclear Physics Past Papers
Solved past paper MCQs for Nuclear Physics from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
Cancerous thyroid is treated with ____. (SZABMU 2024)
A
Cholrine-36
B
Cobalt-60
C
Iodine-131
D
Radium-226
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Nuclear medicine utilizes the natural metabolic behaviors of the human body to deliver targeted radiation.
Solution:
The thyroid gland is the only organ in the body that actively absorbs and stores large quantities of systemic iodine.
By giving a patient Iodine-131 (a strong beta and gamma emitter), the radioactive material is automatically funneled directly into the thyroid.
This intensely localized radiation efficiently kills thyroid cancer cells while minimizing damage to the rest of the body.
Why other options are incorrect:
Cobalt-60 is used for external beam therapy. Chlorine-36 is an environmental tracer. Radium is a heavy bone-seeker that is highly toxic to the whole body.
An atom's identity and mass are governed by the particles inhabiting its nucleus.
Solution:
The nucleus contains protons and neutrons. Together, these heavy particles are collectively referred to as nucleons.
The Mass Number (represented by the symbol \(A\)) is simply the integer sum of the protons and neutrons combined.
Therefore, mass number literally refers to the total number of nucleons in the nucleus.
Why other options are incorrect:
Electrons have negligible mass and don't factor into the mass number. Option D defines the Atomic Number (\(Z\)). Option C represents only a fraction of the mass.
In the context of nuclear physics, universal Greek symbols have highly specific designations.
Solution:
The equation governing the rate of radioactive disintegration is \( \frac{dN}{dt} = -\lambda N \).
In this formula, the symbol \(\lambda\) represents the decay constant.
It physically denotes the probability per unit time that an individual nucleus will spontaneously decay.
Why other options are incorrect:
Dielectric constant uses \(k\) or \(\epsilon_r\). Planck's constant is \(h\). While \(\lambda\) is technically a constant of proportionality in the math, its official physical name is the Decay Constant.
Gamma ray camera can observe radiations from the ____ that are concentrated in the organs: (KU 2024)
A
Atoms
B
Isotopes
C
Nucleons
D
Neutrons
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Correct Key: Option BDiagnostic Explanation
Concept:
A Gamma camera is a specialized diagnostic imaging tool used in nuclear medicine to visualize metabolic functions inside the human body.
Solution:
Doctors inject patients with specific radioactive isotopes (like Technetium-99m) attached to pharmaceutical compounds.
These isotopes selectively accumulate in specific target organs (like bones, heart, or kidneys) based on the body's chemistry.
As the isotopes decay, they emit gamma rays from inside the organ, which pass through the body and are captured by the Gamma camera to form a map of the organ's function.
Why other options are incorrect:
Although isotopes are made of atoms/nucleons, the specific medical term for the active radioactive tracers introduced into the patient is "radioactive isotopes."
If we have "\(N_o\)" number of any radioactive element then after a period of "n" half-lives the number of atoms left behind is (UHS 2024)
A
\( 2^n N_o \)
B
\( \left(\frac{1}{2}\right)^n N_o \)
C
\( (1/2 N_o)^n \)
D
\( (2 N_o)^n \)
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Correct Key: Option BDiagnostic Explanation
Concept:
This is the fundamental universal mathematical equation for exponential radioactive decay based on discrete half-life intervals.
Solution:
After 1 half-life, you have \( \frac{1}{2} \) of the initial amount.
After 2 half-lives, you have \( \frac{1}{2} \times \frac{1}{2} = \left(\frac{1}{2}\right)^2 \) of the initial amount.
Generalizing this pattern, after \(n\) half-lives, the remaining amount is strictly multiplied by the fraction \( \left(\frac{1}{2}\right)^n \).
Yielding the standard formula: $$ N = N_0 \left(\frac{1}{2}\right)^n $$
Why other options are incorrect:
Option A describes exponential growth (like bacterial replication), not decay. Options C and D apply the exponent incorrectly to the initial quantity (\(N_o\)) itself, which makes zero mathematical sense.
Which of the following is NOT the Somatic biological effect of radiation? (UHS 2024)
A
Skin burn
B
Loss of hair
C
Induction of cancer
D
Genes mutation
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Correct Key: Option DDiagnostic Explanation
Concept:
Biological effects of radiation are divided into Somatic (affecting the direct victim) and Genetic (affecting descendants).
Solution:
"Somatic" refers to standard body cells. Damage here results in localized burns, hair loss, cataracts, and eventually cellular cancer.
Gene mutation in reproductive germ cells (sperm and ova) does not harm the exposed individual directly. Instead, it creates inheritable deformities in future generations.
Therefore, gene mutation is strictly classified as a Genetic effect, not a Somatic effect.
Why other options are incorrect:
Skin burns, loss of hair, and somatic cancer are all textbook examples of direct Somatic damage.
An artificial radioactive element can be made by bombarding (UHS 2024)
A
High energy particles on unstable elements
B
Low energy particles on unstable elements
C
High energy particles on stable elements
D
Low energy particles on stable elements
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Artificial (induced) radioactivity occurs when humanity intentionally creates radioactive isotopes that do not naturally exist on Earth.
Solution:
To create a new radioactive element, you must start with a naturally available, stable element as the target material.
You then fire high energy particles (like protons, neutrons, or alpha particles from a particle accelerator or reactor) into the stable nucleus.
The extreme kinetic energy overcomes nuclear repulsive forces, forcing the nucleus to absorb the particle, changing its proton/neutron ratio, and rendering it unstable (radioactive).
Why other options are incorrect:
Bombarding an already unstable element is redundant for generating entirely new artificial isotopes from scratch. Low energy particles usually bounce off the target due to severe Coulombic repulsion.
This means 1 rad represents the absorption of a tiny amount of energy (0.01 Joules) per kilogram of flesh.
Why other options are incorrect:
Option B flips the units upside down. Options C and D only include single dimensions (energy or mass) instead of a ratio, invalidating them as measurements of absorbed dose.
Radiation trauma to human biology is clinically broken down into somatic (body) and genetic (reproductive) damage.
Solution:
Somatic effects destroy the living tissues directly struck by the radiation beam.
Because skin cells are rapidly dividing and exist on the outermost surface of the body, high doses of radiation cause intense cellular death on the surface.
This manifests rapidly as severe erythema, physically presenting as skin burns.
Why other options are incorrect:
Gene and chromosomal deformations passed down via reproductive cells are genetic effects. While "eye burns" (cataracts) are possible, "skin burns" is the universally cited, primary acute somatic effect.
Each person experiences the background radiation dose in one year: (NUMS 2024)
A
1 mSv
B
1 mGy
C
1 Gy
D
1 Sv
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Correct Key: Option ADiagnostic Explanation
Concept:
Natural background radiation emanates constantly from cosmic rays, radon gas, and trace radioactive elements in soil and food.
Solution:
While global averages vary from 1 to 3 mSv depending on altitude and local geology, a baseline figure of approximately 1 mSv (milliSievert) per year is the universally taught textbook standard for public exposure minimums.
Because Sieverts (Sv) account for the biological impact of different radiations (Equivalent Dose), it is the correct unit for measuring human health risk over time.
Why other options are incorrect:
1 Gy or 1 Sv are massive, catastrophic acute radiation doses that would induce severe radiation sickness. 1 mGy only measures raw energy absorbed, ignoring biological hazard weightings.
Which of the following statements is correct: I. nuclear radiation with least penetrating power has most ionization power II. nuclear radiation with most penetrating power has least ionization power (BUMHS 2024)
A
I
B
II
C
Both I and II
D
Neither I nor II
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
There is a direct inverse relationship between a particle's ability to ionize matter and its ability to penetrate matter.
Solution:
Statement I: Alpha particles are bulky and heavily charged (+2). They crash into countless electrons instantly, giving them the most ionization power. Because they lose all their energy doing this, they stop quickly, giving them the least penetrating power. (Statement I is True).
Statement II: Gamma rays are massless and uncharged. They fly through matter mostly undetected, giving them the least ionization power. Because they don't lose energy hitting electrons, they travel very deep, giving them the most penetrating power. (Statement II is True).
Why other options are incorrect:
Because both physical principles are highly accurate textbook definitions, selecting only one ignores the validity of the other.
Mass number of an atom represents the number of: (BUMHS 2024)
A
Proton
B
Neutron
C
Neutron plus proton
D
Proton plus electron
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Correct Key: Option CDiagnostic Explanation
Concept:
The mass of an atom is overwhelmingly concentrated purely within its nucleus.
Solution:
Protons and Neutrons each weigh approximately 1 atomic mass unit (amu).
Electrons are nearly 2000 times lighter, contributing almost zero significant mass to the atom.
Therefore, the total Mass Number (\(A\)) is calculated by adding together the total count of heavy nucleons: Neutrons plus Protons.
Why other options are incorrect:
Protons alone are the Atomic Number (\(Z\)). Neutrons alone are simply the neutron number (\(N\)). Adding electrons provides a mathematically meaningless value.
Let T is the half-life of certain radioactive element and \(N_0\) are the number of atoms present in the sample at \(t = 0\). After time 3T, what percent of atoms present at \(t = 0\) will have been decayed? (BUMHS 2024)
A
12.5%
B
50%
C
87.5%
D
100%
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Correct Key: Option CDiagnostic Explanation
Concept:
You must distinguish carefully between the percentage of atoms remaining and the percentage of atoms that have decayed.
A time of \(3T\) means exactly \(n = 3\) half-lives have passed.
Calculate the remaining fraction: $$ \left(\frac{1}{2}\right)^3 = \frac{1}{8} $$
Convert \(1/8\) to a percentage: \( \frac{1}{8} = 12.5\% \) of the atoms remain.
The question asks what percent decayed: $$ 100\% - 12.5\% = 87.5\% $$
Why other options are incorrect:
12.5% is the amount remaining, not decayed (this is a very common trap). 50% is the amount decayed after 1 half-life. Radioactive substances never reach 100% mathematical decay in finite time.
Which of the following isotope of hydrogen is unstable? (BUMHS 2024)
A
\(\text{H}^1\)
B
\(\text{D}^2\)
C
\(\text{T}^3\)
D
All of these
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Correct Key: Option CDiagnostic Explanation
Concept:
Hydrogen possesses three naturally occurring isotopes, differentiated strictly by the number of neutrons in the nucleus.
Solution:
Protium (\(\text{H}^1\)): Contains 1 proton, 0 neutrons. It is highly stable and makes up 99.98% of the universe's hydrogen.
Deuterium (\(\text{D}^2\)): Contains 1 proton, 1 neutron. It is completely stable and is found in heavy water.
Tritium (\(\text{T}^3\)): Contains 1 proton and 2 neutrons. Because the neutron-to-proton ratio (2:1) is too high for such a light nucleus, the strong nuclear force cannot maintain stability.
Consequently, Tritium is unstable and undergoes beta decay with a half-life of 12.3 years.
Why other options are incorrect:
Protium and Deuterium are unconditionally stable. Only Tritium is radioactive.
Understanding the distinction between physical dose and biological effect.
Solution:
Gray (Gy) measures the absolute physical energy absorbed per kg of tissue (Absorbed Dose).
Sievert (Sv) measures the Equivalent Dose, which factors in the biological damage caused by different types of radiation (e.g., Alpha is far more damaging per Gray than Gamma).
Therefore, the strictly defined SI unit for Equivalent Dose is the Sievert.
Why other options are incorrect:
Gray is absorbed dose. Rad is an old, non-SI unit for absorbed dose. Mass is a fundamental property, not a radiation unit.
The decay rate of radioactive substance is: (UHS 2023)
A
Constant with time
B
Varies inversely with time
C
Decrease exponentially
D
Decreases linearly with time
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The rate of decay (Activity) depends strictly on the number of unstable nuclei present at any given moment.
Formula:
$$ A = A_0 e^{-\lambda t} $$
Solution:
Because nuclei are constantly decaying, the total number of remaining unstable nuclei continuously drops.
As the pool of available nuclei drops, there are fewer atoms available to decay in the next second.
This creates a mathematical curve where the decay rate decreases exponentially as time progresses.
Why other options are incorrect:
If it were constant, elements would vanish instantly at the end of their lifespan. Inverse and linear functions do not model the probabilistic nature of quantum decay.
The Curie (Ci) is a standardized, historical unit of radioactivity derived from the behavior of Radium.
Solution:
Originally, 1 Curie was defined as the decay rate of exactly one gram of Radium-226.
Through modern physical measurement, this value has been rigidly standardized to exactly \( 3.7 \times 10^{10} \text{ disintegrations per second} \).
This massive number shows why measuring activity in Becquerels (1 decay/s) often requires prefixes like Mega or Giga.
Why other options are incorrect:
Option A (7.3) is a fabricated distractor. Options C and D completely miss the massive exponent (\(10^{10}\)) necessary for macroscopic radiation amounts.
The rate of radioactive decay is directly proportional to the stability of the: (SZABMU 2023)
A
Medium
B
Atmosphere
C
Isotopes
D
Half life
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The fundamental law of radioactive decay connects the decay rate strictly to the properties of the specific unstable atoms.
Formula:
$$ \frac{\Delta N}{\Delta t} = -\lambda N $$
Solution:
The formula states that the rate of decay is directly proportional to \(N\), which is the number of active, unstable isotopes currently present in the sample.
The rate has absolutely nothing to do with external conditions like atmospheric pressure or surrounding medium.
is slightly awkward, but "isotopes" is the only option representing the physical nuclei undergoing decay.
Why other options are incorrect:
Decay is fully independent of macroscopic medium and atmosphere. It is inversely proportional to half-life, not directly proportional.
Skin burn, loss of hair, drop in white blood cell and induction of cancer are example of ____ of radiation. (SZABMU 2023)
A
Zener effect
B
Genetic effect
C
Somatic effect
D
Biological effect
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Correct Key: Option CDiagnostic Explanation
Concept:
Radiation effects in humans are divided clinically into Somatic effects and Genetic effects.
Solution:
Somatic effects manifest directly in the body of the individual who was exposed to the radiation.
Because radiation destroys actively dividing cells, tissues like skin, hair follicles, and bone marrow (which makes white blood cells) are damaged first, leading to burns, hair loss, and immune suppression.
Cancer induction is also a delayed somatic mutation.
Why other options are incorrect:
Genetic effects refer to mutations passed to offspring. Zener effect is a semiconductor electronics concept. While "Biological effect" is technically true, "Somatic effect" is the required specific scientific classification.
Radiations are dangerous to living organism because they damage the cell by: (SINDH 2023)
A
By producing ions in cells
B
By increasing the temperature of cells
C
By decreasing the number of cells
D
By destroying the cells
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Nuclear radiation damages biological tissue primarily through a microscopic, atomic-level mechanism.
Solution:
When high-energy particles (alpha, beta, gamma) crash into the water molecules and DNA inside human cells, they carry enough energy to knock electrons out of orbit.
This massive creation of ions and highly reactive free radicals severely breaks molecular bonds.
These broken bonds corrupt DNA chains, leading directly to cell death or uncontrolled cancerous growth.
Why other options are incorrect:
While cell death (Option D and C) is the result, Option A explains the fundamental physical mechanism of the damage. Radiation heating (Option B) is negligible compared to the ionization trauma.
Because one proton is lost but exactly one neutron is instantly created, the total count of nucleons (protons + neutrons) inside the nucleus does not change.
Therefore, the mass number (nucleon number, \(A\)) is strictly conserved.
Note: This conservation of mass number holds true for all forms of beta decay (beta-plus, beta-minus, and electron capture).
Why other options are incorrect:
Option B violates the fundamental conservation laws of nuclear physics. Options C and D do not grammatically answer the status of a numerical quantity.
Radon-222 is a naturally occurring radioactive noble gas that often seeps into basements and poses an environmental health hazard.
Solution:
It is a standard physical constant that the half-life of the most stable isotope of Radon (\(^{222}\text{Rn}\)) is approximately 3.8 days.
This specific timeframe allows the gas enough time to diffuse out of soil and rock before decaying into solid, highly toxic radioactive Polonium isotopes inside human lungs.
Why other options are incorrect:
These are strictly distractor values. 3.8 minutes is too fast; 3.8 months/years is too long for this specific isotope.
The [time] taken for half the number of atoms of radioactive isotopes to disintegration is called: (NUMS 2023)
A
Average life
B
Mean life
C
Total life
D
Half life
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
This question relies entirely on recognizing a fundamental definition in nuclear physics.
Solution:
The exact interval of time required for 50% of the radioactive nuclei in a specific sample to undergo radioactive decay is defined universally as its Half-life (\(T_{1/2}\)).
Why other options are incorrect:
Average (Mean) life is defined as \(1/\lambda\), which is roughly 1.44 times longer than the half-life. Total life is essentially infinite.
A low energy neutron has RBE of 10. How much energy is absorbed by a man of mass 80 kg if the value of equivalent dose is 400 rem? (UHS 2022)
A
16 J
B
32 J
C
48 J
D
64 J
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Correct Key: Option BDiagnostic Explanation
Concept:
Equivalent Dose (\(D_e\)) measures the biological effect of radiation, calculated by multiplying the Physical Dose (\(D\)) by the Relative Biological Effectiveness (RBE). Physical Dose is the absorbed energy per unit mass.
It has been observed that Thorium (\(_{90}^{234}\text{Th}\)) is transformed into Protactinium (\(_{91}^{234}\text{Pa}\)) after the emission of ____ particle: (UHS 2022)
A
Alpha
B
Beta
C
Gamma
D
Alpha, Beta, Gamma
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Correct Key: Option BDiagnostic Explanation
Concept:
Identify the emitted particle by applying the conservation of mass number (\(A\)) and atomic number (\(Z\)).
Solution:
Parent nucleus: \(_{90}^{234}\text{Th}\).
Daughter nucleus: \(_{91}^{234}\text{Pa}\).
Change in Mass Number (\(A\)): \(234 - 234 = 0\).
Change in Atomic Number (\(Z\)): \(91 - 90 = +1\).
A particle with mass number 0 and charge number -1 is an electron (\(_{-1}^{0}\text{e}\)), which corresponds to a Beta particle.
Why other options are incorrect:
Alpha emission would decrease mass by 4. Gamma emission would not change either number.
The half-life of Strontium (Sr) is 8.70 hours. Its decay constant is: (UHS 2022)
A
0.000022 s
B
45000 /s
C
0.000022 /s
D
0.000032 /s
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The decay constant (\(\lambda\)) is inversely proportional to the half-life. Time must be strictly converted into SI units (seconds) to obtain a standard decay constant.
Option A has the wrong unit (seconds instead of per second). Option B is a massive math error (multiplying instead of dividing). Option D uses incorrect numbers.
Iodine-131 is a critical and heavily tested medical radioisotope utilized primarily for treating and diagnosing thyroid disorders.
Solution:
It is a standard factual constant that the physical half-life of Iodine-131 is exactly 8.02 days (commonly rounded to 8 days).
This specific timeframe is biologically ideal—it lives long enough to reach the thyroid and deliver a therapeutic dose, but decays rapidly enough to prevent permanent whole-body radiation toxicity.
Why other options are incorrect:
These are distractor figures. For reference, Strontium-89 is roughly 50 days, not Iodine.
Skin burns, loss of hair, drop in the white blood cells etc are examples of: (SZABMU 2022)
A
Somatic effect
B
Genetic effect
C
Metabolism effect
D
Mutation effect
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Radiation biological damage is categorized into two main types: Somatic (affecting the exposed individual directly) and Genetic (affecting future offspring).
Solution:
"Soma" means body. Somatic effects occur when radiation damages regular body cells (like skin, hair follicles, and bone marrow).
Symptoms like localized burns, epilation (hair loss), and leukopenia (drop in white blood cells) appear directly in the irradiated person.
Genetic effects only occur when reproductive germ cells (sperm/egg) are mutated, passing defects to the next generation.
Why other options are incorrect:
Genetic and mutation effects relate to inherited DNA damage passed to children. "Metabolism effect" is not a recognized standalone category of acute radiation syndrome.
The half-life of a radio-active sample predicts about: (ETEA 2022)
A
Whole life of sample
B
Disintegration time of half number of atoms
C
Decay only
D
Total time for stable atoms
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Correct Key: Option BDiagnostic Explanation
Concept:
This question tests the strict definition of radioactive half-life.
Solution:
Half-life (\(T_{1/2}\)) is defined as the exact interval of time required for half of the unstable parent nuclei in a given sample to undergo radioactive disintegration.
Because decay is statistical, you can never predict exactly which atom will decay, but you can confidently predict that 50% of the entire population will transform over this time period.
Why other options are incorrect:
A radioactive sample technically takes an infinite time to decay completely, so "whole life" is meaningless. Option C is too vague. Option D does not define the rate at which they stabilize.
Which radiation cannot be generated under electron transitions in different orbits? (ETEA 2022)
A
Infrared
B
Ultraviolet
C
X-rays
D
\(\gamma\)-rays
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Correct Key: Option DDiagnostic Explanation
Concept:
Electromagnetic radiation is emitted from different parts of the atom depending on the underlying physical process.
Solution:
Infrared, visible, ultraviolet, and X-rays are all produced by electrons shifting between different atomic orbitals (energy shells) outside the nucleus.
Gamma (\(\gamma\)) rays are strictly a nuclear phenomenon. They are emitted exclusively when protons and neutrons inside the nucleus transition from a higher excited energy state to a lower, stable state.
Therefore, orbital electron transitions can never produce gamma rays.
Why other options are incorrect:
Infrared (Paschen/Brackett series), UV (Lyman series), and X-rays (K-shell transitions) are all fundamentally driven by orbital electron physics.
Radioactivity is an internal property of atomic nuclei, governed by the strong and weak nuclear forces, which operate on enormous energy scales.
Solution:
The rate of decay (Activity, \(A\)) depends mathematically on the initial number of atoms (\(N_0\)), the nature of the material (decay constant \(\lambda\)), and the passage of time (\(t\)).
Macroscopic environmental factors like temperature, pressure, and chemical bonding possess energy levels in the electron-volts (eV) range, which are utterly insufficient to penetrate or influence the nucleus (MeV range).
Therefore, heating or freezing a radioactive material has zero effect on its half-life or decay rate.
Why other options are incorrect:
Activity explicitly depends on \(N_0\), material type, and time elapsed via the formula \(A = A_0 e^{-\lambda t}\).
The Curie (Ci) is a traditional, non-SI unit of measurement in nuclear physics.
Solution:
It was originally defined as the radioactivity (decay rate) of exactly 1 gram of Radium-226.
It is precisely standardized today as \(3.7 \times 10^{10}\) disintegrations per second.
Therefore, it solely measures the activity of a radioactive source.
Why other options are incorrect:
Temperature is Kelvin/Celsius. Half-life is measured in seconds/years. Magnetism uses Tesla/Weber. (Note: "Curie Temperature" is a concept in magnetism, but "Curie" alone is the unit of activity).
Which one is stable element in the followings? (ETEA 2022)
A
Lead
B
Plutonium
C
Radium
D
Protactinium
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Correct Key: Option ADiagnostic Explanation
Concept:
Elements with atomic numbers \(Z > 82\) are naturally radioactive and decay continuously until they reach a stable nuclear configuration.
Solution:
Plutonium (\(Z=94\)), Radium (\(Z=88\)), and Protactinium (\(Z=91\)) are all heavy, highly unstable radioactive elements.
Lead (Pb), with an atomic number of \(Z = 82\), is the heaviest stable element. The three major natural radioactive decay series (Uranium, Actinium, Thorium) all terminate with stable, non-radioactive isotopes of Lead.
Why other options are incorrect:
Plutonium, Radium, and Protactinium lack any stable isotopes and exist in a state of continuous decay.
For the treatment of cancer, the source of gamma rays used, is: (ETEA 2022)
A
Co-60
B
Iodine-126
C
Na-15
D
Pb-207
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Correct Key: Option ADiagnostic Explanation
Concept:
External beam radiotherapy requires an intense, reliable source of high-energy gamma photons to destroy deep-seated tumors.
Solution:
Cobalt-60 (Co-60) is a synthetic radioisotope that undergoes beta decay followed immediately by the emission of two highly penetrating gamma rays (1.17 MeV and 1.33 MeV).
Because of its strong gamma emission and convenient half-life (~5.27 years), it is the global gold standard in "Gamma Knife" surgeries and external radiation therapy.
Why other options are incorrect:
Iodine is used internally for the thyroid. Lead-207 is totally stable. Na-15 is an extreme, instantly decaying isotope not used in medicine.
Radio isotopes can be formed by bombardment with ____ particles (BMU 2022)
A
Neutron
B
Photons
C
Electrons
D
Positron
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Correct Key: Option ADiagnostic Explanation
Concept:
Artificial radioactivity is induced by striking a stable nucleus with a high-energy particle, forcing it to absorb the particle and become an unstable isotope.
Solution:
Protons, electrons, and alpha particles all carry electrical charge. If fired at an atom, they are powerfully repelled by the target's electron cloud or the positive nucleus due to electrostatic forces.
A Neutron carries no electrical charge. It faces zero Coulombic repulsion and can effortlessly penetrate deep into the target nucleus, even at very low speeds.
Once absorbed, it alters the neutron-to-proton ratio, transforming the stable atom into a radioactive isotope.
Why other options are incorrect:
Charged particles (electrons, positrons) are heavily deflected. Photons lack rest mass and cannot alter the nucleon count to create new isotopes.
Which radiations cause more ionization? (MDCAT 2021)
A
Alpha
B
Beta
C
Gamma
D
None
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Ionizing power depends directly on the mass and electrical charge of a particle. Particles with higher charge and mass interact more strongly with surrounding atoms.
Solution:
Alpha (\(\alpha\)) particles consist of two protons and two neutrons, giving them a heavy mass of 4 amu and a double positive charge (\(+2e\)).
Because they are massive and heavily charged, they move relatively slowly and exert a massive electromagnetic pull on atomic electrons.
This causes them to aggressively strip electrons from nearby atoms, making them the most ionizing type of nuclear radiation.
Why other options are incorrect:
Beta particles have only a \(-1e\) charge and tiny mass, making them moderately ionizing. Gamma rays are uncharged, massless photons, making them the least ionizing.
Beta decay is a nuclear transmutation process. In standard terminology, "beta rays" default to beta-minus (\(\beta^{-}\)) particles unless specified otherwise.
Solution:
During \(\beta^{-}\) decay, a neutron in the nucleus converts into a proton, emitting a fast-moving particle to conserve charge.
This ejected particle has a charge of \(-1e\) and a mass equivalent to an electron.
Therefore, standard beta rays are simply streams of highly energetic electrons originating from the nucleus.
Why other options are incorrect:
While positrons are emitted in \(\beta^{+}\) decay, the historical and standard use of "beta ray" refers to the electron. Protons are heavy nucleons, not beta particles.
\( 4.5 \times 10^9 \) year is the half-life of: (NUMS 2020)
A
\(\text{U}^{238}\)
B
\(\text{U}^{235}\)
C
\(\text{U}^{236}\)
D
\(\text{U}^{234}\)
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Correct Key: Option ADiagnostic Explanation
Concept:
This is a standard memorization question regarding the primordial isotopes crucial for Earth's radiometric dating.
Solution:
Uranium-238 (\(\text{U}^{238}\)) is the most abundant natural isotope of Uranium and has a highly stable, immensely long half-life of \(4.468 \times 10^9\) years (often rounded to \(4.5 \times 10^9\) in textbooks).
This duration closely mirrors the physical age of the Earth, making U-238 essential in geological age-dating.
Why other options are incorrect:
U-235 has a much shorter half-life of about 700 million years. U-234 has a half-life of 245,000 years.
In relation \(\lambda T_{1/2} = 0.693\), which quantity is represented by \(\lambda\). (MDCAT 2019)
A
Half-life
B
Activity
C
Wavelength
D
Decay constant
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Correct Key: Option DDiagnostic Explanation
Concept:
This question requires knowledge of standard nuclear physics notation and definitions.
Solution:
The equation given is the standard mathematical relation between half-life and decay probability.
\( T_{1/2} \) stands for half-life.
\( \lambda \) mathematically represents the decay constant, defining the probability of an individual nucleus decaying per unit time.
Why other options are incorrect:
Although \(\lambda\) is the universal symbol for wavelength in optics and waves, in nuclear decay equations, it universally stands for the decay constant. Activity is represented by \(A\).
The main difference between X-Rays and \(\gamma\)-Rays is: (ETEA 2019)
A
Frequency
B
Wave length
C
Energy
D
Origin
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Both X-rays and Gamma rays are high-energy electromagnetic waves. Their properties (energy, frequency, wavelength) overlap heavily in the spectrum, making them physically identical photons at equal energies.
Solution:
The strict scientific distinction between them is purely their origin (source).
Gamma (\(\gamma\)) rays originate deep inside the nucleus as it transitions from an excited state to a stable state.
X-rays originate completely outside the nucleus, generated by the deceleration of fast electrons or transitions of orbital electrons between atomic shells.
Why other options are incorrect:
Because their energy spectra overlap (you can have hard X-rays more energetic than soft Gamma rays), energy, frequency, and wavelength are not the absolute defining differences.
While using radiation therapy, cancerous thyroid is treated with ____ radioisotope: (ETEA 2019)
A
Carbon
B
\(^{235}\text{uranium}\)
C
Thorium
D
Iodine-131
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Medical radioisotopes are selected based on the specific metabolic properties of organs in the human body.
Solution:
The human thyroid gland naturally and actively absorbs and concentrates circulating iodine to synthesize thyroid hormones.
By administering the radioactive isotope Iodine-131, the thyroid rapidly absorbs it.
The I-131 then undergoes beta decay directly inside the tumor, emitting highly localized radiation that destroys the cancerous thyroid cells without heavily damaging distant tissues.
Why other options are incorrect:
Carbon-14 is used for radiocarbon dating. Uranium and Thorium are highly toxic heavy metals used for nuclear power and strictly not used in human bodies.
Calculate the activity (decaying atom per unit time) of radioactive strontium-90 having \( 6.7 \times 10^{21} \) atoms at t=0 decay constant of strontium-90 is \( 8.3 \times 10^{-10} \text{ s}^{-1} \)? (MDCAT 2018)
A
\( 8.01 \times 10^{10} \text{ Bq} \)
B
\( 5.6 \times 10^{12} \text{ Bq} \)
C
\( 5.6 \times 10^{11} \text{ s}^{-1} \)
D
\( 5.6 \times 10^{10} \text{ Bq} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Activity (\(A\)) is defined as the rate at which nuclei undergo decay. It is mathematically the product of the decay constant and the number of active nuclei.
Rounding to two significant figures yields \( 5.6 \times 10^{12} \text{ Bq} \).
Why other options are incorrect:
Options A and D represent errors in shifting the decimal place (exponent math). Option C uses the wrong units (Activity is measured in Becquerels, Bq, although it is equivalent to \(s^{-1}\), conventionally Bq is used for activity, and the magnitude \(10^{11}\) is an unshifted decimal error).
Two radioactive samples, \(S_1\) and \(S_2\) have half-live 3 hours and 7 hours respectively. If they have the same activity at certain instant t, what is the ratio of the number of atoms of \(S_1\) to \(S_2\) at instant t? (ETEA 2018)
A
9 : 49
B
49 : 9
C
3 : 7
D
7 : 3
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Activity (A) is defined as \( A = \lambda N \). If two samples have equal activities, their quantities must be inversely proportional to their decay constants.
Gamma (\(\gamma\)) radiation is the emission of a massless, uncharged high-energy photon.
Solution:
Because a photon has a mass number (\(A\)) of 0 and a charge number (\(Z\)) of 0, emitting it does not remove any protons or neutrons from the nucleus.
The parent nucleus \(_{Z}^{A}\text{X}\) transitions to a lower energy state but retains its identity.
Therefore, the daughter nucleus Y remains identical in nucleon composition: \(_{Z}^{A}\text{Y}\).
Why other options are incorrect:
Option A shows alpha decay. Option B shows beta minus decay. Option C shows beta plus decay or electron capture.
Which of the following has maximum ionizing power? (MDCAT 2017)
A
\(\alpha\)
B
\(\gamma\)
C
\(\beta\)
D
Neutron
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Correct Key: Option ADiagnostic Explanation
Concept:
Ionizing power depends on a particle's electrical charge and mass, which govern how forcefully it interacts with electrons in target atoms.
Solution:
The alpha (\(\alpha\)) particle contains 2 protons, giving it a strong +2e charge, and has a large mass of 4 amu.
Because of its large mass, it travels relatively slowly, spending more time near atomic electron clouds. Combined with its +2e charge, it rapidly and violently pulls electrons away from nearby atoms, causing intense ionization over a very short distance.
Therefore, \(\alpha\) particles have the highest ionizing power, significantly greater than beta or gamma rays.
Why other options are incorrect:
Gamma rays have no charge and weak ionization. Beta particles have -1e charge and low mass, giving moderate ionization. Neutrons have no charge, ionizing indirectly via nuclear collisions.
A radioactive nucleus X undergoes a series of decay according to the scheme: $$ \text{X} \xrightarrow{\alpha} \text{X}_1 \xrightarrow{\beta} \text{X}_2 \xrightarrow{\alpha} \text{X}_3 \xrightarrow{\gamma} \text{X}_4 $$ If the mass number and atomic number of X are 180 and 72 respectively, the corresponding number of \(\text{X}_4\) are: (MDCAT 2017)
A
176, 69
B
172, 69
C
176, 71
D
172, 71
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Trace the distinct changes to mass number (A) and atomic number (Z) step-by-step. Alpha (\(\alpha\)) subtracts 4 from A, 2 from Z. Beta (\(\beta\)) changes A by 0, adds 1 to Z. Gamma (\(\gamma\)) changes nothing.
Solution:
Start: X (A=180, Z=72)
Step 1 (\(\alpha\)): to \(\text{X}_1\) A = 180 - 4 = 176 Z = 72 - 2 = 70
Step 2 (\(\beta\)): to \(\text{X}_2\) A = 176 - 0 = 176 Z = 70 + 1 = 71
Step 3 (\(\alpha\)): to \(\text{X}_3\) A = 176 - 4 = 172 Z = 71 - 2 = 69
Step 4 (\(\gamma\)): to \(\text{X}_4\) A = 172 - 0 = 172 Z = 69 - 0 = 69
Final Result: Mass number 172, Atomic number 69.
Why other options are incorrect:
They represent math errors or incomplete steps. Option A misses the second alpha decay. Options C and D miss the atomic number decrease from the second alpha decay.
On the electromagnetic spectrum, energy and frequency are directly proportional, while energy and wavelength are inversely proportional (\(E = \frac{hc}{\lambda}\)).
Solution:
Gamma (\(\gamma\)) rays originate from nuclear transitions and carry the highest energy in the entire electromagnetic spectrum.
X-rays originate from inner electron orbital transitions and possess slightly less energy than gamma rays.
Because \(\text{Energy}_{\gamma} > \text{Energy}_{\text{X-ray}}\), it mathematically follows that \(\lambda_{\gamma} < \lambda_{\text{X-ray}}\). Thus, gamma rays have a shorter wavelength.
Why other options are incorrect:
Option C asserts gamma rays have less energy. Option A is false (they occupy distinct, albeit overlapping, energy bands based on origin). Option D is nonsensical terminology for wavelength.
Emission of \(\gamma\)-rays from radioactive element result into: (MCAT 2016)
A
Increase of charge number by 1
B
No change in the charge number
C
Decrease of mass number by 1
D
Decrease charge number by 1
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Gamma rays are highly energetic photons of electromagnetic radiation, not physical matter.
Solution:
A photon has no rest mass and carries no electrical charge.
When a nucleus relaxes from an excited state (\(\text{X}^{*}\)) and emits a gamma ray, it only releases excess energy.
Therefore, there is absolutely no change to the number of protons (charge number/atomic number \(Z\)) or total nucleons (mass number \(A\)).
Why other options are incorrect:
Option A describes beta decay. Options C and D describe physically non-standard or alternative decay modes (like positron emission) completely unrelated to pure gamma emission.
The relation between decay constant '\(\lambda\)' and half-life '\(T_{1/2}\)' of radioactive substance is: (MCAT 2016)
A
\( \lambda = \frac{1}{T_{1/2}} \)
B
\( \lambda = T_{1/2} \)
C
\( \lambda = 0.693 \, T_{1/2} \)
D
\( \lambda = \frac{0.693}{T_{1/2}} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The half-life of a radioactive isotope is mathematically derived from the radioactive decay law \(N = N_0 e^{-\lambda t}\).
Formula:
$$ T_{1/2} = \frac{\ln(2)}{\lambda} $$
Solution:
The natural logarithm of 2 (\(\ln 2\)) is approximately 0.693.
Substitute this into the formula: $$ T_{1/2} = \frac{0.693}{\lambda} $$
Rearrange algebraically to solve for \(\lambda\): $$ \lambda = \frac{0.693}{T_{1/2}} $$
Why other options are incorrect:
Option A lacks the \(\ln 2\) constant. Option B incorrectly implies they are equal. Option C proposes a direct proportionality, but they are inversely proportional.
Radiation damages living organism is primarily due to: (ETEA 2016)
A
Excitation phenomena
B
Ionization
C
Photo electric effect
D
Heating
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Correct Key: Option BDiagnostic Explanation
Concept:
Nuclear radiation (alpha, beta, gamma) carries immense kinetic energy compared to molecular bonds.
Solution:
When these high-energy particles pass through biological tissue, they violently strip electrons off cellular atoms, a process known as ionization.
This ionization breaks molecular bonds, corrupts DNA, and produces highly reactive free radicals inside the cell, which directly causes radiation sickness, mutations, and cancer.
Why other options are incorrect:
While excitation and photoelectric effect occur, they are secondary sub-mechanisms. Heating from standard environmental radiation is negligible. Ionization is the primary destructive mechanism.
In the reaction $$ _{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + \text{e} $$, the electron \(\text{e}\) emits from the (MCAT 2015)
A
\(1^{\text{st}}\) orbit
B
Nucleus
C
\(2^{\text{nd}}\) orbit
D
Valence shell
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
This equation represents a nuclear beta minus (\(\beta^{-}\)) decay, where Thorium-234 transmutes into Protactinium-234.
Solution:
Beta decay is strictly a nuclear process mediated by the weak force.
A neutron deep inside the nucleus transforms into a proton, creating an electron and an antineutrino in the process.
This highly energetic electron is immediately ejected directly from the nucleus, not from the atomic orbital electron clouds.
Why other options are incorrect:
Options A, C, and D refer to atomic orbitals (electron shells). Electrons ejected from there constitute ionization (via photoelectric effect or Auger effect), not nuclear transmutation.
According to the equation $$ _{Z}^{A}\text{X} \rightarrow \text{Y} + 3\alpha \text{ particles} $$, what are the atomic and mass numbers of 'Y'? (MCAT 2015)
A
Z-6, A-12
B
Z+1, A
C
Z-2, A-4
D
Z+3, A
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The principle of conservation of nucleon number and charge number dictates that the sums of \(A\) and \(Z\) on both sides of a nuclear equation must be perfectly equal.
Solution:
The decay emits 3 alpha particles. One alpha particle is \(_{2}^{4}\text{He}\).
Total mass number removed by 3 alphas: \(3 \times 4 = 12\).
Total atomic number removed by 3 alphas: \(3 \times 2 = 6\).
Set up the conservation equation: $$ _{Z}^{A}\text{X} \rightarrow _{Z_Y}^{A_Y}\text{Y} + 3 \, (_{2}^{4}\text{He}) $$
Solve for Y: $$ A_Y = A - 12 $$ $$ Z_Y = Z - 6 $$
Why other options are incorrect:
Option C is the result for only 1 alpha particle. Options B and D describe beta or other physically incorrect decay paths.
A certain radioactive nuclide of mass number 'x' decay by \(\beta\)-emission and \(\alpha\)-emission to a second nuclide of mass number 't', which of the following correctly relates 'x' and 't'? (MCAT 2015)
A
x = t - 4
B
x + 3 = t
C
x = t + 4
D
x - 1 = t
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
We must track the changes purely to the mass number during sequential decay.
Solution:
Initial mass number = \(x\).
Beta Emission: A neutron turns into a proton. Mass number does not change. So, \(\Delta A_{\beta} = 0\).
Alpha Emission: The nucleus ejects a helium nucleus (2p + 2n). Mass number decreases by 4. So, \(\Delta A_{\alpha} = -4\).
The final mass number \(t\) is the initial mass \(x\) minus 4: $$ t = x - 4 $$
Rearranging the equation to solve for \(x\): $$ x = t + 4 $$
Why other options are incorrect:
Option A sets \(t = x + 4\), implying the nucleus gained mass, which is impossible in decay. Options B and D introduce incorrect constants.
During the decay of radioactive isotopes \( _{90}^{232}\text{Th} \) to a stable isotope, six \(\alpha\)-particles and four \(\beta\)-particles are emitted, what is the atomic number 'Z' and mass number 'A' of the stable isotopes: (MCAT 2015)
A
Z = 70, A = 220
B
Z = 82, A = 212
C
Z = 78, A = 212
D
Z = 82, A = 208
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Correct Key: Option DDiagnostic Explanation
Concept:
Calculate the total cumulative changes to both the atomic number (\(Z\)) and mass number (\(A\)) caused by multiple decay events.
Three points of radioactive radiation are observed as shown in the figure presence of electric field, which type of radiation is shown in the path '1'? (MCAT 2014)
Deflection of Radiations (1, 2, 3) in Electric Field
A
Alpha
B
Beta
C
Gamma
D
Cathode ray
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
When radiations pass between electrically charged plates, they deflect according to Coulomb's Law: opposite charges attract.
Solution:
In standard test diagrams of this type, a radioactive source emits three distinct beams.
The beam deflecting strongly toward the positive plate must be negatively charged. Beta particles are high-speed electrons (negative), thus they bend toward the positive plate.
(Note: Gamma travels straight through un-deflected. Alpha deflects slightly toward the negative plate due to its positive charge and heavy mass).
Why other options are incorrect:
Alpha is positive. Gamma is neutral. Cathode rays are electrons but aren't classified as radioactive nuclear radiation.
Because the nucleus gains one new proton, its Atomic Number (\(Z\)) increases by 1.
Because a neutron is simply exchanged for a proton, the total count of nucleons (protons + neutrons) does not change. Thus, the Mass Number (\(A\)) remains the same.
Why other options are incorrect:
Option D perfectly describes alpha decay. Options A and C propose mass changes that violate fundamental nucleon conservation in beta decay.
A uranium isotope \( _{92}^{234}\text{U} \) undergoes one \(\alpha\)-decay and one \(\beta\)-decay. What is the atomic number of the final product? (MCAT 2014)
A
90
B
89
C
91
D
88
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
We must trace the sequential changes to the atomic number (\(Z\)) caused by each specific decay event.
Solution:
Step 1: Alpha Decay Emitting an alpha particle (\(_{2}^{4}\text{He}\)) subtracts 2 from the atomic number. $$ Z_{\text{intermediate}} = 92 - 2 = 90 $$
Step 2: Beta Decay Emitting a beta particle (\(_{-1}^{0}\text{e}\)) adds 1 to the atomic number. $$ Z_{\text{final}} = 90 + 1 = 91 $$
Why other options are incorrect:
90 is the state after just the alpha decay. 89 would require beta-plus or electron capture. 88 would be the result of two consecutive alpha decays.
A naturally occurring radioactive element decays two alpha particles. Which one of the following represents the status of daughter element with respect to mass number A and charge number Z? (MCAT 2014)
A
Z decreases by 4 and A decreases by 2
B
Z decreases by 4 and A decreases by 8
C
Z decreases by 2 and A decreases by 4
D
Z decreases by 8 and A decreases by 4
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Each individual alpha particle (\(_{2}^{4}\text{He}\)) carries away 2 protons and 4 total nucleons.
Solution:
For two alpha particles emitted simultaneously or sequentially, the total loss must be multiplied by 2.
Change in Charge Number (\(Z\)): $$ \Delta Z = 2 \times (-2) = -4 $$
Change in Mass Number (\(A\)): $$ \Delta A = 2 \times (-4) = -8 $$
Why other options are incorrect:
Option C is the result for only one alpha particle. Option A perfectly reverses the changes of Z and A for a single particle. Option D reverses Z and A for two particles.
The ejected particle removes 2 protons from the parent nucleus. Thus, the atomic number \(Z\) decreases by 2.
The ejected particle removes 4 total nucleons (2 protons + 2 neutrons). Thus, the mass number \(A\) decreases by 4.
Why other options are incorrect:
Option A perfectly swaps the properties of \(Z\) and \(A\). Option B describes positron emission or electron capture. Option C is physically non-standard.
Which one of the following emissions takes place in a nuclear reaction? $$ _{90}\text{Th}^{234} \rightarrow _{91}\text{Pa}^{234} + \_\_\_\_\_\_ $$ (MCAT 2013)
A
Alpha
B
Beta
C
Gamma
D
Photons
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
By analyzing the changes in atomic number (\(Z\)) and mass number (\(A\)) between the parent and daughter nuclei, we can identify the emitted particle.
Solution:
Parent nucleus: Thorium (\(Z=90, A=234\)).
Daughter nucleus: Protactinium (\(Z=91, A=234\)).
The mass number \(A\) remained 234. The atomic number \(Z\) increased from 90 to 91 (\(+1\)).
A process where \(Z\) increases by 1 and \(A\) is unchanged is the emission of an electron, which is a beta (\(\beta^{-}\)) particle.
Why other options are incorrect:
Alpha emission would decrease mass by 4. Gamma emission would not change \(Z\) or \(A\).
Among the three types of radioactive radiation, which have strongest penetration power? (MCAT 2013)
A
Alpha
B
Beta
C
Gamma
D
\(\alpha\), \(\beta\) and \(\gamma\) have same penetration.
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Penetration power refers to a radiation's ability to pass through matter. It is inversely related to the particle's mass, charge, and ionizing capability.
Solution:
Alpha particles are heavy and doubly charged, interacting intensely with matter and stopping within a few centimeters of air or a sheet of paper.
Beta particles are lighter and singly charged, stopping at a thin sheet of aluminum.
Gamma rays are highly energetic, massless, uncharged photons. Because they lack charge, they do not interact strongly with matter and can only be stopped by thick blocks of dense materials like lead or concrete.
Why other options are incorrect:
Alpha has the lowest penetration. Beta is intermediate. They definitely do not have equal penetration power.
Emission of radiation from radioactive substance is (MCAT 2013)
A
Dependent on both temperature and pressure
B
Independent of both temperature and pressure
C
Independent of temperature but dependent on pressure
D
Independent of pressure but dependent on temperature
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Radioactivity is fundamentally a purely nuclear phenomenon, governed by the weak and strong nuclear forces operating deep inside the nucleus.
Solution:
Macroscopic physical conditions such as temperature, pressure, chemical bonding, and electric/magnetic fields only affect the outer electron shells of an atom.
The energy scales of chemical changes (eV) are millions of times weaker than nuclear changes (MeV). Therefore, external environmental factors are completely unable to influence the probability or rate of nuclear decay.
Why other options are incorrect:
Any option asserting that radioactive decay depends on external thermodynamic factors like temperature or pressure is physically incorrect.
What is the charge on alpha particles emitted during the phenomenon of radioactivity? (MCAT 2012)
A
+e
B
-2e
C
-e
D
+2e
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Correct Key: Option DDiagnostic Explanation
Concept:
An alpha particle is physically identical to the nucleus of a helium-4 atom.
Solution:
It contains exactly 2 protons and 2 neutrons.
It contains exactly 0 electrons.
Because each proton carries an elementary positive charge (\(+e\)), the total charge of the 2 protons is \(+2e\).
Why other options are incorrect:
Option A is the charge of a positron or proton. Option B is physically incorrect for standard decay particles. Option C is the charge of an electron/beta particle.
A radioactive nuclide decays by emitting an \(\alpha\)-particle and a \(\gamma\)-ray photon, the change in the nucleon number will be: (MCAT 2012)
A
-4
B
-2
C
+2
D
-3
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The "nucleon number" is another term for the mass number (\(A\)), which is the total count of protons and neutrons in a nucleus.
Solution:
An \(\alpha\)-particle is a helium nucleus (\(_{2}^{4}\text{He}\)), meaning it carries away 4 nucleons (2 protons + 2 neutrons). This decreases the parent's nucleon number by 4.
A \(\gamma\)-ray is a high-energy photon. It carries away excess energy but has zero mass and zero charge, meaning it removes 0 nucleons.
Total change: \(\Delta A = -4 + 0 = -4\).
Why other options are incorrect:
Option B is the change in the atomic number (\(Z\)), not the nucleon number. Options C and D do not match standard decay rules.
A half-life of sodium-24 is ____ which is used to estimate the volume of blood in a patient: (MCAT 2012)
A
6 hours
B
8 hours
C
15 hours
D
15 days
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Sodium-24 (\(^{24}\text{Na}\)) is an artificially produced radioactive isotope commonly used as a medical tracer, specifically to study blood circulation and estimate total blood volume.
Solution:
It is a factual standard that the physical half-life of Sodium-24 is approximately 15 hours.
This timeframe is short enough to minimize prolonged radiation exposure to the patient, but long enough to conduct the medical procedure.
Why other options are incorrect:
These are incorrect memorization distractors. For comparison, Iodine-131 has a half-life of 8 days, and Technetium-99m is roughly 6 hours.
In a radioactive phenomenon, observation shown in figure where \(\alpha\) deviates lesser than \(\beta\) in same electric or magnetic field. What is the reason of less deviation of \(\alpha\)? (MCAT 2012)
Deflection of α, β, and γ Radiations in Electric Field
A
\(\alpha\) is a lighter particle
B
\(\alpha\) is heavier particle
C
\(\alpha\) is very fasting moving particle
D
None of these
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
When charged particles enter a transverse electric or magnetic field, their path is deflected. The radius of curvature (and thus the amount of deviation) depends on the particle's mass, velocity, and charge.
Formula:
$$ r = \frac{mv}{qB} $$
Solution:
Deflection is inversely proportional to mass. A higher mass creates a larger radius of curvature (less visible deviation/bending).
An alpha particle (mass \(\approx 4\text{ amu}\)) is nearly 7,300 times heavier than a beta particle (an electron).
Due to its massive inertia, the alpha particle strongly resists changes in its momentum, causing it to deviate much less than the lightweight beta particle, despite carrying twice the charge.
Why other options are incorrect:
Option A is factually inverted. Option C is false; alpha particles generally travel at \(0.05c\) while beta particles can approach \(0.9c\).
Because a neutron turns into a proton, the number of protons (atomic number \(Z\)) increases by 1.
The total number of nucleons (protons + neutrons) remains exactly the same, so the mass number (\(A\)) is unchanged.
Why other options are incorrect:
Option A is physically impossible (mass cannot increase during natural decay). Option B describes positron (\(\beta^{+}\)) decay or electron capture. Option D is incorrect as mass number doesn't drop by 1.
Ionizing power is the ability of radiation to strip electrons from atoms. It heavily depends on the mass and charge of the particle.
Solution:
Alpha particles have a charge of +2e and a large mass, leading to massive interactions and the highest ionizing power.
Beta particles have a charge of -1e and very small mass, giving them intermediate ionizing power.
Gamma rays are uncharged, massless photons. They interact weakly with matter, giving them the lowest ionizing capability among the three, but the highest penetration power.
Why other options are incorrect:
All other options incorrectly rank the physical properties of the radiations. Gamma is strictly the least ionizing.
The formula shows that \(T_{1/2}\) is equal to a constant divided by \(\lambda\).
Therefore, half-life is mathematically inversely proportional to the decay constant \(\lambda\).
A larger decay constant means a higher probability of decay per second, which results in a much shorter half-life.
Why other options are incorrect:
Option A and C incorrectly involve squares. Option B suggests that a highly active element (large \(\lambda\)) would decay slowly (large \(T_{1/2}\)), which contradicts physics.
The transformation of a neutron into proton in the nucleus gives rise to emission of: (MCAT 2011)
A
Beta particles
B
Gamma particles
C
Alpha particles
D
X-rays
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In unstable, neutron-rich nuclei, the weak interaction can cause a neutron to spontaneously decay into a proton, an electron, and an electron antineutrino.
The newly formed proton remains in the nucleus, increasing the atomic number by 1.
The highly energetic electron (\(_{-1}^{0}\text{e}\)) is immediately ejected from the nucleus. This ejected electron is known as a beta (\(\beta^{-}\)) particle.
Why other options are incorrect:
Gamma emission occurs when an excited nucleus relaxes, without changing protons or neutrons. Alpha decay involves emitting a helium nucleus. X-rays originate from atomic electron transitions.
In the half-life of an element, the equation for the number of decaying atoms is given by: (MCAT 2010)
A
\(\Delta N = N \Delta t\)
B
\(\Delta N \propto -N \Delta t\)
C
\(\Delta N = K N \Delta t\)
D
\(\Delta N = -\lambda N \Delta t\)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The fundamental law of radioactive decay states that the rate of decay of a radioactive sample is directly proportional to the number of radioactive nuclei present at that time.
Formula:
$$ \frac{\Delta N}{\Delta t} = -\lambda N $$
Solution:
\(\Delta N\) is the change in the number of undecayed nuclei.
\(N\) is the current number of nuclei, and \(\Delta t\) is the time interval.
\(\lambda\) is the decay constant.
The negative sign indicates that the number of parent nuclei \(N\) is decreasing over time. Rearranging yields: \(\Delta N = -\lambda N \Delta t\).
Why other options are incorrect:
Option A is missing the decay constant and the negative sign. Option B is a proportionality, not an equation. Option C uses an undefined constant \(K\) and misses the critical negative sign.
Decay constant '\(\lambda\)' is given as: (MCAT 2010)
A
\( -\frac{\Delta N / N}{\Delta t} \)
B
\( -\frac{N}{\Delta t} \)
C
\( -\frac{\Delta N}{\Delta t} \)
D
\( \frac{\Delta N / N}{\Delta t} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The decay constant \(\lambda\) represents the probability of decay per unit time for a given nucleus. It is derived directly from the radioactive decay law.
Solution:
Start with the radioactive decay law: $$ \frac{\Delta N}{\Delta t} = -\lambda N $$
Rearrange the equation to solve for \(\lambda\): $$ \lambda = -\frac{\Delta N}{N \Delta t} = -\frac{\Delta N / N}{\Delta t} $$
The term \(\Delta N / N\) represents the fractional change in the number of nuclei.
Why other options are incorrect:
Option B calculates a frequency, not the decay constant. Option C defines activity (rate of decay), not the decay constant itself. Option D misses the necessary negative sign.
Gamma (\(\gamma\)) radiation is the emission of high-energy electromagnetic photons from a nucleus transitioning from an excited energy state to a lower, more stable state.
Solution:
An excited nucleus is denoted by an asterisk (\(\text{X}^{*}\)).
Because a gamma ray is a photon (mass = 0, charge = 0), emitting it does not change the atomic number (\(Z\)) or the mass number (\(A\)).
The valid representation is: $$ _{Z}\text{X}^{A*} \rightarrow _{Z}\text{X}^{A} + \gamma $$
Why other options are incorrect:
Option A lacks the asterisk denoting the excited parent state. Option B represents a physically incorrect process. Option C incorrectly shows \(Z\) increasing, which occurs during beta decay, not gamma emission.
A certain radioactive mass decays from 64 gm to 2 gm in 20 days. What is its half-life? (MCAT 2008)
A
5 days
B
10 days
C
4 days
D
6 days
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Radioactive decay follows a geometric progression where the remaining amount halves after each half-life period. The formula for the remaining mass is based on the number of half-lives elapsed.
Formula:
$$ N = N_0 \left(\frac{1}{2}\right)^n $$
Where \(n\) is the number of half-lives, defined as \(n = \frac{t}{T_{1/2}}\).
Solution:
Given: Initial mass \(N_0 = 64\text{ g}\), Final mass \(N = 2\text{ g}\), Total time \(t = 20\text{ days}\).
Substitute into the formula: $$ 2 = 64 \left(\frac{1}{2}\right)^n $$
Solve for \(n\): $$ \frac{2}{64} = \frac{1}{32} = \left(\frac{1}{2}\right)^n \implies n = 5 $$
When a helium atom loses an electron, it becomes: (MCAT 2008)
A
An alpha particle
B
A positive helium ion
C
Proton
D
A negative helium ion
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A neutral helium atom consists of 2 protons, 2 neutrons, and 2 orbiting electrons. Ionization is the process of losing or gaining electrons.
Solution:
If a neutral helium atom (\(\text{He}\)) loses exactly one electron, it retains 1 electron and its 2 protons.
Because there is now a net positive charge (+2 from protons, -1 from electron), it becomes a singly ionized positive helium ion (\(\text{He}^{+}\)).
To become an alpha particle, it would need to lose both of its electrons (\(\text{He}^{2+}\)).
Why other options are incorrect:
An alpha particle requires the loss of two electrons. A proton is a hydrogen nucleus, not helium. Losing an electron creates a positive ion, not a negative one.
Beta ray emitted by a radioactive substance is: (MCAT 2008)
A
An electron which was existing outside the nucleus.
B
An electron emitted by the nucleus as a result of the decay of neutron inside the nucleus.
C
An electron which was existing inside the nucleus.
D
A pulse of electromagnetic wave.
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Beta minus (\(\beta^{-}\)) decay occurs when a neutron-rich nucleus transforms a neutron into a proton, emitting an electron and an antineutrino in the process.
Electrons do not inherently "exist" inside the nucleus. They are created at the exact moment of decay.
The weak nuclear force mediates the transformation of a neutron into a proton, generating a high-speed electron (the beta particle) that is immediately ejected.
Why other options are incorrect:
Option A describes ionization, not nuclear radioactivity. Option C is a common misconception; nuclei contain protons and neutrons, not pre-existing electrons. Option D describes gamma rays.
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