Question

An $\alpha $ -particle and a proton are accelerated from rest, by a potential difference of $100 \, V$ . After this, their de-Broglie wavelengths are $\lambda _{\alpha }$ and $\lambda _{p}$ , respectively. The ratio $\frac{\lambda _{p}}{\lambda _{\alpha }}$ , to the nearest integer, is

• A. 3
• B. 2
• C. 4
• D. 5

Answer 1

Answer: (A)

$\frac{1}{2} \text{mv}^{2} = \text{qV}$
$⇒ \, \, \text{p} = \text{mv} = \sqrt{2 \text{mq V}}$
De broglie wavelength,
$\lambda = \frac{\text{h}}{\text{p}} ⇒ \, \, \frac{\lambda _{\alpha }}{\lambda _{\text{p}}} = \frac{\text{P}_{\text{p}}}{\text{P}_{\alpha }} = \sqrt{\frac{\text{m}_{\text{p}} \text{q}_{\text{p}}}{\text{m}_{\alpha } \text{q}_{\alpha }}}$
$⇒ \, \, \frac{\lambda _{\text{p}}}{\lambda _{\alpha }} = \sqrt{8}$
The nearest integer is 3
$\therefore $ answer is 3