Question

A proton accelerated from rest through a potential difference of 'V' volts has a wavelength $\lambda$ associated with it. An alpha particle in order to have the same wavelength must be accelerated from rest through a potential difference of:

• A. V volt
• B. 4V volt
• C. 2V volt
• D. $\frac{V}{8}volt$

Answer 1

The correct option is D $\frac{V}{8}volt$

$\lambda =\frac{h}{\sqrt{2meV}}$ -------(i)

$\lambda =\frac{h}{\sqrt{2\left(4m\right)\left(2e\right)V^{{}^{\prime }}}}$ -------(ii)

Equating equation (i) & (ii) we have

$2meV=2\left(4m\right)\left(2e\right)V^{{}^{\prime }}$

So $V^{{}^{\prime }}=\frac{V}{8}.$

Hence, the correct option is $D$