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

Vedclass · Accepted Answer

$\frac{\lambda}{\sqrt{2}}$

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(D) The de-Broglie wavelength $\lambda$ of an electron accelerated by a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Let $\lambda_1$ be the initial wavelength at voltage $V_1 = 10 \ kV$ and $\lambda_2$ be the final wavelength at voltage $V_2 = 20 \ kV$.Then,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$.Substituting the given values: $\frac{\lambda_2}{\lambda} = \sqrt{\frac{10 \ kV}{20 \ kV}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Therefore,$\lambda_2 = \frac{\lambda}{\sqrt{2}}$.

An electron beam,when accelerated by a voltage of $10 \ kV$,has a de-Broglie wavelength of $\lambda$. If the voltage is increased to $20 \ kV$,then the de-Broglie wavelength associated with the electron beam would be:

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