Question

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

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

Answer 1

The correct option is D $2\sqrt{2}:1$

De Broglie wavelength is given by:

$\lambda =\frac{h}{p}$

Writing momentum as a a function of kinetic energy and mass

$\lambda =\frac{h}{\sqrt{2Em}}$

and Kinetic energy = potential through which it is accelerated times the charge on it

$\lambda =\frac{h}{\sqrt{2Vqm}}$
$\frac{{\lambda }_{proton}}{{\lambda }_{alpha}}=\sqrt{\frac{q_{alpha}{m}_{alpha}}{q_{proton}{m}_{proton}}}$

$q_{alpha}=4e$
$q_{proton}=e$
$m_{alpha}=4m_{proton}$

Hence, $\frac{{\lambda }_{proton}}{{\lambda }_{alpha}}=\sqrt{\frac{8}{1}}=2\sqrt{2}$