The energy that should be added to an electro | vedclass

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(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$,where $E$ is the kinetic energy of the electron.This implies $\la

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Thrice the initial energy

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(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$,where $E$ is the kinetic energy of the electron.This implies $\lambda \propto \frac{1}{\sqrt{E}}$,or $E \propto \frac{1}{\lambda^2}$.Let the initial wavelength be $\lambda_1 = 10^{-10} \ m$ and the final wavelength be $\lambda_2 = 0.5 	imes 10^{-10} \ m$.Then,$\frac{E_2}{E_1} = \left( \frac{\lambda_1}{\lambda_2} ight)^2 = \left( \frac{10^{-10}}{0.5 	imes 10^{-10}} ight)^2 = (2)^2 = 4$.Thus,the final energy $E_2 = 4E_1$.The energy that should be added is $\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$.

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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