Q118 / 155 · WBJEE 2019 · Physics
A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $\left(\lambda_{p}\right)$ and that of electron $\left(\lambda_{c}\right) ?$
- A$\lambda_{p}=2000 \lambda_{e}$
- B$\lambda_{p}=\frac{\lambda_{e}}{2000}$
- C$\lambda_{p}=20 \sqrt{5} \lambda_{0}$
- Correct answerD$\lambda_{p}=\frac{\lambda_{0}}{20 \sqrt{5}}$
Solution
As we know that de-Broglie wavelength is given as $\lambda=\frac{h}{p}=\frac{h}{m v}$
where, $h=$ Planck constant
$$
\begin{array}{l}
p=\text { momentum of particle } \\
v=\text { velocity of particle }
\end{array}
$$
and $m=$ mass of the particle.
Eq.(i) can be written as,
$$
\lambda=\frac{h}{2 m(\mathrm{KE})}=\frac{h}{\sqrt{2 m q v}} \quad[\because \mathrm{KE}=q v]
$$
where, $\mathrm{KE}=$ Kinetic energy of particle
Hence, $\lambda \propto \frac{1}{\sqrt{m}}$
$$
\text { Now, } \frac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=\sqrt{\frac{m_{\text {electron }}}{m_{\text {proton }}}}
$$
Given, mass of proton, $m_{\text {proton }}=2000 \mathrm{m}_{\text {electron }}$
$\frac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=\sqrt{\frac{1}{2000}}$
$\frac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=\frac{1}{20 \sqrt{5}} \Rightarrow \lambda_{p}=\frac{\lambda_{e}}{20 \sqrt{5}}$
Hence, the correct answer is option: 'D'.