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(B) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $K$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.This implies $\lambda \pro

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thrice the initial energy

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(B) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $K$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.This implies $\lambda \propto \frac{1}{\sqrt{K}}$,or $K \propto \frac{1}{\lambda^2}$.Let $K_1$ be the initial energy corresponding to $\lambda_1 = 1 \ nm$.Let $K_2$ be the final energy corresponding to $\lambda_2 = 0.5 \ nm$.Then,$\frac{K_2}{K_1} = \left( \frac{\lambda_1}{\lambda_2} \right)^2 = \left( \frac{1 \ nm}{0.5 \ nm} \right)^2 = (2)^2 = 4$.So,$K_2 = 4K_1$.The additional energy required is $\Delta K = K_2 - K_1 = 4K_1 - K_1 = 3K_1$.Thus,the additional energy is thrice the initial energy.

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

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