Question

A hydrogen atom of mass m is at rest makes a transition from its first excited state to ground state. As a result, a photon of wavelength $\lambda$ is emitted and the hydrogen atom recoils. If $E_0$ is the ionization energy of a hydrogen atom, the Plancks constant is h and c is the speed of light in vacuum, the recoil speed is.

• A. $\frac{3E_0}{4mc}$
• B. $\frac{3E_0}{mc}$
• C. $\frac{h}{2m\lambda }$
• D. $\frac{h}{m\lambda }$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{3E_0}{4mc}$
D $\frac{h}{m\lambda }$

$\text{From conservation of momentum}$

$\begin{array}{c}mv=\frac{hf}{c}=\frac{hc}{\lambda c}=\frac{h}{\lambda }\\ v=\frac{h}{m\lambda }\end{array}$

$\begin{array}{c}\text{Assuming all the energy is taken}\\ \text{by the photon, the transition}\\ \text{energy for the transition from}\\ \text{its first excited state to}\\ \text{ground state of the hydrogen}\\ \text{atom is :-}hf=\Delta E=E_0(\frac{1}{1^2}-\frac{1}{2^2})\end{array}$

$\begin{array}{cc}\Rightarrow & hf=\Delta E=\frac{3E_0}{4}\\ & \Rightarrow hf=\frac{3E_0}{4}\\ V& =\frac{hf}{cm}=\frac{3E_0}{4mc}=\frac{h}{m\lambda }\end{array}$