Question

An electron acelerated under a potential difference of $V\text{(in V)}$ has a certain wavelength $\lambda .$ Mass of the proton is $2000$ times the mass of an electron. If the proton has to have the same wavelength $\lambda$, then it will have to be accelerated under potential difference of $\text{(in V)}$

• A. $\frac{V}{\sqrt{2000}}$
• B. $\frac{V}{2000}$
• C. $\frac{V}{\sqrt{1000}}$
• D. $\frac{V}{1000}$

Answer 1

The correct option is B $\frac{V}{2000}$

de-Broglie wavelength is,

$\lambda =\frac{h}{p}=\frac{h}{mv}$

$KE=qV$

$\frac{p^2}{2m}=qV\Rightarrow p=\sqrt{2mqV}$

$\lambda =\frac{h}{\sqrt{2mqV}}$

For, ${\lambda }_e={\lambda }_p$

$\Rightarrow \frac{h}{\sqrt{2m_e{q}_{e}V}}=\frac{h}{\sqrt{2m_p{q}_p{V}_p}}$

$\Rightarrow V_P=\frac{m_e{q}_{e}V}{m_p{q}_p}[\because q_e=q_p]$

$V_p=\frac{V}{2000}[\because m_p=2000m_e]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.