Physics - Modern Physics Question with Solution | TestHub

The de-Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda =\frac{h}{\sqrt{2meV}}$.

This implies $\lambda \propto \frac{1}{\sqrt{V}}$, or $V\propto \frac{1}{{\lambda }^2}$.

For the first case, $V_1\propto \frac{1}{{\lambda }^2}$.

For the second case, $V_2\propto \frac{1}{(2\lambda {)}^2}=\frac{1}{4{\lambda }^2}$.

Taking the ratio of the two potentials:

$\frac{V_1}{V_2}=\frac{\frac{1}{{\lambda }^2}}{\frac{1}{4{\lambda }^2}}=4$

Thus, $\frac{V_1}{V_2}=4:1$.

Answer: $4:1$