Question

An $\alpha$ -particle and a proton are accelerated from rest, by a potential difference of $100V$ . 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)

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