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

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
$\Delta E_{2 \rightarrow 1}=E\left(1 - \frac{1}{2^{2}}\right)=\frac{3 E}{4}$
$\frac{3 E}{4}=\frac{1}{2}mv^{2}+\frac{h c}{\lambda }$
$\frac{3 E}{4}=\frac{1}{2}mv^{2}+mvc$
$\left(\frac{m}{2}\right)v^{2}+\left(m c\right)v-\left(\frac{3 E}{4}\right)=0$
$\Rightarrow v=\left[\sqrt{\frac{3 E}{2 m} + c^{2}}\right]-c$