Question

A proton and an alpha particle are subjected to same potential difference $V$. Their de-Broglies wavelengths ${\lambda }_p,{\lambda }_{\alpha }$ will be in the ratio.

• A. $2:1$
• B. $2\sqrt{2}:1$
• C. $4:1$
• D. $1:2$

Answer 1

Answer: (B)

$2\sqrt{2}:1$

Let the mass of alpha be $4m$ and charge be $2e$

therefore mass of proton is $m$ and charge is $e$

Energy of proton $E_P=e\times V$

So momentum is $P_P=\sqrt{2Em}=\sqrt{2eVm}$

Similar for alpha $P_{\alpha }=\sqrt{2\times 2e\times V\times 4m}$

${\lambda }_p=\frac{h}{P_p}$

${\lambda }_{\alpha }=\frac{h}{P_{\alpha }}$

$\frac{{\lambda }_{\alpha }}{{\lambda }_P}=2\sqrt{2}$