The energy that should be added to an electro | vedclass

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(B) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $E$ by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.From this,we see that $\l

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Thrice the initial energy

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(B) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $E$ by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.From this,we 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 $\frac{10^{-10}}{0.5 	imes 10^{-10}} = \sqrt{\frac{E_2}{E_1}}$.This simplifies to $2 = \sqrt{\frac{E_2}{E_1}}$,so $\frac{E_2}{E_1} = 4$,or $E_2 = 4E_1$.The energy that must be added is $\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$.Thus,the added energy is thrice the initial energy.

The energy that should be added to an electron,to reduce its de-Broglie wavelength from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$,will be

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