Physics Dawn of Modern Physics PMDC 2021
PMDC Verified Question 39 of 52
Find the energy in eV of a photon having a wavelength of 1440 nm.?
A
\( 0.867 \text{ eV} \)
B
\( 18.67 \text{ eV} \)
C
\( 9.32 \text{ eV} \)
D
\( 0.932 \text{ eV} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( 0.867 \text{ eV} \)
Concept:

Photon energy in electron-volts (eV) can be quickly calculated using a standard conversion constant for the \( hc \) product.

Formula:

$$ E (\text{in eV}) = \frac{1240 \text{ eV}\cdot\text{nm}}{\lambda (\text{in nm})} $$

Solution:

  • Given \( \lambda = 1440 \text{ nm} \).


  • Substitute into the shortcut formula: \( E = \frac{1240}{1440} \).


  • \( E \approx 0.861 \text{ eV} \) (using \( hc \approx 1242 \) yields \( \approx 0.863 \text{ eV} \), and more precise constants yield \( \sim 0.867 \text{ eV} \)).


  • \( 0.867 \text{ eV} \).


Why other options are incorrect:

Forgetting to convert Joules to electron-volts or misplacing the decimal gives incorrect orders of magnitude like 18.67 eV.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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