Question

A charged particle is accelerated from rest through a certain potential difference. The de Broglie wavelength is ${\lambda }_1$ when it is accelerated through $V_1$ and is ${\lambda }_2$ when accelerated through $V_2$. The ratio ${\lambda }_1/{\lambda }_2$ is

• A. $V_1^{3/2}:V_2^{3/2}$
• B. $V_2^{1/2}:V_1^{1/2}$
• C. $V_1^{\frac{1}{2}}:V_2^{\frac{1}{2}}$
• D. $V_1^2:V_2$

Answer 1

Answer: (B)

$V_2^{1/2}:V_1^{1/2}$

de Broglie wavelength is given by $\lambda =\frac{h}{p}$
where momentum $p=\sqrt{2mK}$ and kinetic energy $K=qV$
$⟹\lambda =\frac{h}{\sqrt{2mqV}}$
We get $\lambda \propto \frac{1}{\sqrt{V}}$
$⟹{\lambda }_1:{\lambda }_2=V_2^{1/2}:V_1^{1/2}$