Atoms And Nuclei Question 438
Question: The following diagram indicates the energy levels of a certain atom when the system moves from 2E level to E, a photon of wavelength $ \lambda $ is emitted. The wavelength of photon produced during its transition from $ \frac{4E}{3} $ level to E is
Options:
A) $ \lambda /3 $
B) $ 3\lambda /4 $
C) $ 4\lambda /3 $
D) $ 3\lambda $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ 2E-E=\frac{hc}{\lambda }\Rightarrow E=\frac{hc}{\lambda } $ $ \frac{4E}{3}-E=\frac{hc}{\lambda ‘}\Rightarrow \frac{E}{3}=\frac{hc}{\lambda ‘}$
$\therefore \frac{\lambda ‘}{\lambda }=3\Rightarrow \lambda ‘=3\lambda $
BETA
