Physics Atomic Spectra KU 2024
PMDC Verified Question 3 of 21
If an electron in the hydrogen atom jumps from second to first orbit, the emitted radiation has a wavelength of?
A
4/3RH
B
3/4RH
C
RH
D
4RH
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: 4/3RH
Concept: Wavelength of emitted radiation is mathematically derived from the Rydberg formula based on initial and final principal quantum numbers.

Formula:
$$ \frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) $$

Solution:
  • The electron transitions from the second orbit (\( n = 2 \)) to the first orbit (\( p = 1 \)).
  • Substitute the values: \( \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \).
  • Simplify the parentheses: \( \frac{1}{\lambda} = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \).
  • To find \( \lambda \), invert the equation: \( \lambda = \frac{4}{3R_H} \).


Why other options are incorrect:
Option B (\( \frac{3}{4}R_H \)) is the un-inverted wave number (\( 1/\lambda \)) rather than wavelength. Options C and D involve incorrect algebraic simplifications. Note: The source text formats the fraction exactly as 4/3RH.

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