Question

A proton and an alpha particle are accelerating by the same potential difference. Find the ratio of their de-Broglie wavelength? $\left(\text{charge}\right(q_{\alpha })=+2e,q_{proton}=+e\text{and}{m}_{\alpha }=4m_{proton})$

• A. $\sqrt{16}$
• B. $\sqrt{6}$
• C. $\sqrt{8}$
• D. $\sqrt{2}$

Answer 1

The correct option is C $\sqrt{8}$

As the proton and alpha particle accelerating by the same potential difference. Hence, their KE will be

$KE=qV$

$\Rightarrow \frac{1}{2}\frac{p^2}{m}=qV$

$\Rightarrow p=\sqrt{2mqV}[\because V=\text{same for both}]$

$\therefore \frac{p_p}{p_{\alpha }}=\sqrt{\frac{q_p{m}_p}{q_{\alpha }{m}_{\alpha }}}$

de-Broglie wavelength is given by,

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

$\Rightarrow \lambda \propto \frac{1}{p}$

$\therefore \frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{p_{\alpha }}{p_p}=\sqrt{\frac{q_{\alpha }{m}_{\alpha }}{q_p{m}_p}}$

$\because q_{\alpha }=+2e,q_p=+e\text{and}{m}_{\alpha }=4m_p$

$\therefore \frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{(\frac{+2e}{+e}\times \frac{4m_p}{m_p})}=\sqrt{8}$

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

Why this question? Key note: Keep remember charge of alpha particle, $q_{\alpha }=+2e$. Sometimes it is not given in problem.