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(B) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.From this,we have $E = \fr

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$3$

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(B) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.From this,we have $E = \frac{h^2}{2m\lambda^2}$,which means $E \propto \frac{1}{\lambda^2}$.Let the initial energy be $E_1$ corresponding to wavelength $\lambda$. So,$E_1 = \frac{k}{\lambda^2}$ (where $k = \frac{h^2}{2m}$).The final wavelength is $\lambda_2 = \frac{\lambda}{2}$. The final energy $E_2$ is $E_2 = \frac{k}{(\lambda/2)^2} = \frac{4k}{\lambda^2} = 4E_1$.The energy to be added is $\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$.Given that $\Delta E = nE_1$,we find $n = 3$.

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

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