Question

The following diagram indicates the energy levels of a certain atom. When the system moves from energy level $2E$ to $E$ , a photon of wavelength $\lambda$ is emitted. The wavelength of photon produced during its transition from energy level $\frac{4E}{3}$ to $E$, 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 $3\lambda$

For the transition, $2E\to E$
$2E-E=\frac{hc}{\lambda }$
$\Rightarrow E=\frac{hc}{\lambda }.......1$
For the transition, $\frac{4E}{3}\to E$
$\frac{4E}{3}-E=\frac{hc}{{\lambda }^{{}^{\prime }}}$
$\Rightarrow \frac{E}{3}=\frac{hc}{{\lambda }^{{}^{\prime }}}....2$

Dividing $1$ by $2$, we get,
$\frac{E}{\frac{E}{3}}=\frac{\frac{hc}{\lambda }}{\frac{hc}{{\lambda }^{{}^{\prime }}}}$
$3=\frac{{\lambda }^{{}^{\prime }}}{\lambda }$
$\Rightarrow {\lambda }^{{}^{\prime }}=3\lambda$

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

Key idea: $\Delta E=\frac{hc}{\lambda }$ $\Delta E$ is the change in the energy $h$ is plank's constant $\lambda$ is wavelength $c$ is velocity of light in vacuum.