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 emitted during its transition from $\frac{4E}{3}$ level to $E$ level is:

• A. $\frac{\lambda }{3}$
• B. $\frac{3\lambda }{4}$
• C. $\frac{4\lambda }{3}$
• D. $3\lambda$

Answer 1

The correct option is D $3\lambda$

When the system moves from $2E$ level to $E$ level, then,

$\Delta E=2E-E=\frac{hc}{\lambda }$

$\Rightarrow E=\frac{hc}{\lambda }....\left(1\right)$

When the system moves from $\frac{4}{3}E$ level to $E$ level, then,

$\Delta E=\frac{4}{3}E-E=\frac{hc}{{\lambda }^{\prime }}$

$\Rightarrow \frac{E}{3}=\frac{hc}{{\lambda }^{\prime }}....\left(2\right)$

Dividing equation $\left(1\right)$ by equation $\left(2\right)$, we get,

$\frac{E}{E/3}=\frac{hc}{\lambda }\times \frac{{\lambda }^{\prime }}{hc}$

$\Rightarrow 3=\frac{{\lambda }^{\prime }}{\lambda }$

$\Rightarrow {\lambda }^{\prime }=3\lambda$

Hence, option $\left(D\right)$ is correct.

Key Idea: $\Delta E=\frac{hc}{\lambda }$ Where, $\lambda$ is the wavelength of radiation and $\Delta E$ is the change in the energy from one state to another.