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(A) The de-Broglie wavelength $\lambda$ of an electron with energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.From this relation,we can see that

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$3 E$

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(A) The de-Broglie wavelength $\lambda$ of an electron with energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$,which implies $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{E_2}{E_1}}$.Given $\lambda_1 = 10^{-10} \ m$ and $\lambda_2 = 0.5 	imes 10^{-10} \ m$,we have $\frac{\lambda_1}{\lambda_2} = \frac{10^{-10}}{0.5 	imes 10^{-10}} = 2$.Substituting this into the ratio equation: $2 = \sqrt{\frac{E_2}{E_1}}$.Squaring both sides,we get $4 = \frac{E_2}{E_1}$,so $E_2 = 4E_1 = 4E$.The energy imparted to the electron is $\Delta E = E_2 - E_1 = 4E - E = 3E$.

How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$? (Let $E$ be the initial energy of the electron).

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