Question

When a hydrogen atom emits a photon during a transition from $n=4$ to $n=2$, its recoil speed is about

• A. $4.28m{s}^{-1}$
• B. $0.814m{s}^{-1}$
• C. $2.07m{s}^{-1}$
• D. $0.407m{s}^{-1}$

Answer 1

Answer: (B)

We know that momentum of a photon, $p=\frac{h}{\lambda }$
$\Rightarrow mv=\frac{h}{\lambda }\Rightarrow v=\frac{h}{m\lambda }$..(i)
$[\because p=mv]$
Given, hydrogen atom emits a photon during a transition from $n=4$ to $n=2$, According to Bohr's formula for $H$ -atom, the wavelength of the emitted photon is given by
$\therefore \frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
where, $R=$ Rydberg's constant $=1.097\times 10^7{m}^{-1}$
$\Rightarrow \frac{1}{\lambda }=1.097\times 10^7\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$
$\Rightarrow \lambda =4.86\times 10^{-7}m$
$\because$ Mass of a proton $=1.6\times 10^{-27}kg$
Now, from Eq. (i), we get
Hence, recoil speed of photon,
$v=\frac{6.62\times 10^{-34}}{1.6\times 10^{-27}\times 4.86\times 10^{-7}}$
$=0.8513m/s$
or $v\approx 0.814m{s}^{-1}$