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

Answer 1

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

We know, $\lambda$ = $\frac{h}{\sqrt{2meV}}$
where $m$ = mass of the particle
e = chrage on the particle
V = Potential difference
So, $\lambda$ for proton = ${\lambda }_p$ = $\frac{h}{\sqrt{2m_{p}eV}}$ ..(i)
mass of alpha particle = 4 times mass of proton
charge on alpha particle = 2 times charge on proton
So, $\lambda$ for alpha particle = ${\lambda }_{\alpha }$ = $\frac{h}{\sqrt{2\left(4m_p\right)\left(2e\right)V_{\alpha }}}$ ...(ii)
On comparing, equation (i) and (ii) ,
$V_{\alpha }=$$\frac{V}{8}$