Question

A proton and an $ \alpha $ -particle are accelerated through same voltage. The ratio of their de - Broglie wavelength will be

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

Answer 1

Answer: (B)

Here, $q V=\frac{1}{2} m v^{2}$
or $m v=\sqrt{2 q m V}$
So, de-Broglie wavelength,
$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 q m V}}$
i.e, $\lambda \propto \frac{1}{\sqrt{q m}}$
Hence, $\frac{\lambda_{p}}{\lambda_{a}}=\sqrt{\frac{2 e \times 4 m}{e \times m}}=2 \sqrt{2}$