Question

A Proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is

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

Answer 1

Answer: (C)

$2\sqrt{2}$

Answer 2

Key Point The de-Broglie wavelength of a particle of mass $m$ and moving with velocity $v$ is given by $\lambda=\frac{h}{m v} (\because p=m v)$ de-Broglie wavelength of a proton of mass $m_{1}$ and kinetic energy $k$ is given by $\lambda_{1} =\frac{h}{\sqrt{2 m_{1} k}} (\because p=\sqrt{2 m k})$ $=\frac{h}{\sqrt{2 m_{1} q v}} \ldots \text { (i) }[\because k=q V]$ For an alpha particle mass $m_{2}$ carrying charge $q_{0}$ is accelerated through potential $V$, then $\lambda_{2}=\frac{h}{\sqrt{2 m_{2} q_{0} V}}$ $\because$ For $\alpha$ -particle $\left({ }_{2}^{4} He \right)$ $\therefore q_{0}=2 q$ and $m_{2}=4 m_{1}$ $\therefore \lambda_{2}=\frac{h}{\sqrt{2 \times 4 m_{1} \times 2 q \times V}}$ ...(ii) The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get $\frac{\lambda_{1}}{\lambda_{2}} =\frac{h}{\sqrt{2 m_{1} q V}} \times \frac{\sqrt{2 \times m_{1} \times 4 \times 2 q V}}{h}= \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$ $=2 \sqrt{2}$