The diagram shows the energy levels of a certain atom. When the system moves from level 2E to level E, a photon of wavelength \( \lambda \) is emitted. The wavelength of the photon produced in the transition from level \( 4E/3 \) to level E is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 3\lambda \)
When an atom moves from a higher level to a lower one, the energy difference is carried away by a photon, \( hc/\lambda=E_i-E_f \). So the wavelength is inversely proportional to the energy difference.
From the figure, the levels are E, 4E/3 and 2E. For the transition from 2E to E:
\( \begin{aligned} \frac{hc}{\lambda} &= 2E-E \\ &= E \end{aligned} \)
For the transition from 4E/3 to E, with wavelength \( \lambda' \):
\( \begin{aligned} \frac{hc}{\lambda'} &= \frac{4E}{3}-E \\ &= \frac{E}{3} \end{aligned} \)
Dividing the first result by the second:
\( \begin{aligned} \frac{\lambda'}{\lambda} &= \frac{E}{E/3} \\ &= 3 \end{aligned} \)
\( \lambda'=3\lambda \)