Q197 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
The energy levels of certain atom are represented in the following figure. During the transition from 2E to E level, a photon of wavelength $\lambda$ is emitted. The wavelength of photon produced during transition from $\dfrac{4E}{3}$ level to E will be
- Correct answerA$3\lambda$
- B$\dfrac{\lambda}{3}$
- C$\dfrac{3\lambda}{2}$
- D$\dfrac{4\lambda}{3}$
Solution
The energy of an emitted photon during a transition is given by $\Delta E = \dfrac{hc}{\lambda}$.
For the transition from $2E$ to $E$:
$\Delta E_1 = 2E - E = E$
$\Rightarrow E = \dfrac{hc}{\lambda}$
For the transition from $\dfrac{4}{3}E$ to $E$, let the wavelength be $\lambda'$:
$\Delta E_2 = \dfrac{4}{3}E - E = \dfrac{E}{3}$
$\Rightarrow \dfrac{E}{3} = \dfrac{hc}{\lambda'}$
Substituting $E = \dfrac{hc}{\lambda}$ into the above equation:
$\dfrac{1}{3} \left( \dfrac{hc}{\lambda} \right) = \dfrac{hc}{\lambda'}$
$\Rightarrow \lambda' = 3\lambda$
Answer: $3\lambda$
For the transition from $2E$ to $E$:
$\Delta E_1 = 2E - E = E$
$\Rightarrow E = \dfrac{hc}{\lambda}$
For the transition from $\dfrac{4}{3}E$ to $E$, let the wavelength be $\lambda'$:
$\Delta E_2 = \dfrac{4}{3}E - E = \dfrac{E}{3}$
$\Rightarrow \dfrac{E}{3} = \dfrac{hc}{\lambda'}$
Substituting $E = \dfrac{hc}{\lambda}$ into the above equation:
$\dfrac{1}{3} \left( \dfrac{hc}{\lambda} \right) = \dfrac{hc}{\lambda'}$
$\Rightarrow \lambda' = 3\lambda$
Answer: $3\lambda$
Hence, the correct answer is option: 'A'.