Question

A changed 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. $V_1^2:V_2^2$

Answer 1

The correct option is B $V_2^{1/2}:V_1^{1/2}$

If a charged particle $q$ is accelerated through a potential difference $V$ , then energy acquired by particle is ,

$E=qV$ ....................eq1

Now , the de-Broglie wavelength of a particle of mass $m$ is given by ,

$\lambda =\frac{h}{\sqrt{2mE}}$ ..............eq2

putting the value of $E$ in eq2 from eq1 , we get

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

Now , when charged particle $q$ is accelerated through a potential difference $V_1$ , then

${\lambda }_1=\frac{h}{\sqrt{2mq{V}_1}}$ ..........eq3

When charged particle $q$ is accelerated through a potential difference $V_1$ , then

${\lambda }_2=\frac{h}{\sqrt{2mq{V}_2}}$ ..............eq4

Dividing eq 3 by eq4 , we get

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{{\sqrt{V}}_2}{{\sqrt{V}}_1}=\frac{V_2^{1/2}}{V_1^{1/2}}$