Question

Calculate the ratio of wavelength for an $\alpha $ particle and proton accelerated through same potential difference.

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

Answer 1

Answer: (C)

Using de Broglie equation $\lambda =\frac{h}{\sqrt{2 q m V}}$

$\lambda =$ wavelength, q = charge, m = mass, V = Potential and h = Plank constant

$\lambda _{a}=\frac{h}{\sqrt{2 q_{a} m_{a} V}}$ $\lambda _{p}=\frac{h}{\sqrt{2 q_{p} m_{p} V}}$

$m_{a}=4m_{p},q_{a}=2q_{p}$

$\frac{\lambda _{a}}{\lambda _{p}}=\sqrt{\frac{q_{p} m_{p}}{q_{a} m_{a}}}=\sqrt{\frac{1}{2} . \frac{1}{4}}=\sqrt{\frac{1}{8}}$

$\frac{\lambda _{a}}{\lambda _{p}}=\frac{1}{2 \sqrt{2}}$