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

Vedclass · Accepted Answer

$2 \lambda$

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(D) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Let the initial wavelength be $\lambda_1 = 4\lambda$ at potential $V_1 = V$.Let the final wavelength be $\lambda_2$ at potential $V_2 = 4V$.Using the proportionality $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$,we get:$\frac{\lambda_2}{4\lambda} = \sqrt{\frac{V}{4V}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$\lambda_2 = 4\lambda 	imes \frac{1}{2} = 2\lambda$.

When an electron is accelerated through a potential $V$,the de-Broglie wavelength associated with it is $4 \lambda$. When the accelerating potential is increased to $4V$,its wavelength will be

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