Question

A stationary hydrogen atom in the first excited state emits a photon. If the mass of the hydrogen atom is $m$ and its ionization energy is $E$ , then the recoil velocity acquired by the atom is [speed of light = $c$ ]

• A. $\left[\sqrt{\frac{3E}{2m}+c^2}\right]-c$
• B. $\left[\sqrt{\frac{3E}{4m}+c^2}\right]-c$
• C. $\frac{3E}{4mc}$
• D. $\frac{E}{mc}$

Answer 1

Answer: (A)

$mv=\frac{h}{\lambda }$ , here $v$ is recoil speed of $H$ -atom and $\lambda$ is the wavelength of the photon.
When the electron jumps down from the first excited state to the ground state the energy released is
$\Delta E_{2\to 1}=E\left(1-\frac{1}{2^2}\right)=\frac{3E}{4}$
$\frac{3E}{4}=\frac{1}{2}m{v}^2+\frac{hc}{\lambda }$
$\frac{3E}{4}=\frac{1}{2}m{v}^2+mvc$
$\left(\frac{m}{2}\right)v^2+\left(mc\right)v-\left(\frac{3E}{4}\right)=0$
$\Rightarrow v=\left[\sqrt{\frac{3E}{2m}+c^2}\right]-c$