Question

A hydrogen atom of mass ${}^{\prime }m{}^{\prime }$ being at rest, makes a transition from its first excited state to ground state. As a result a photon of wavelength ${}^{\prime }\lambda {}^{\prime }$ is emitted and the hydrogen atom recoils. If ${}^{\prime }{E}_0{}^{\prime }$ is the ionization energy of a hydrogen atom, the Planck's 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)

$\frac{3E_0}{4mc}$
$\frac{h}{m\lambda }$

From conservation of momentum
$mv=\frac{hf}{c}$
Assuming all the energy is taken by the photon, the transition energy for the transition from its first excited state to ground state of the hydrogen atom is: $hf=\Delta E=\frac{3}{4}{E}_0$

$⟹v=\frac{3E_0}{4mc}=\frac{h}{m\lambda }$