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(D) We know that the de-Broglie wavelength $\lambda$ is related to kinetic energy $K$ by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.This implies th

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three-times the initial energy

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(D) We know that the de-Broglie wavelength $\lambda$ is related to kinetic energy $K$ by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.This implies that $K \propto \frac{1}{\lambda^2}$.Let $K_1$ be the initial kinetic energy at $\lambda_1 = 1 \,nm$ and $K_2$ be the final kinetic energy at $\lambda_2 = 0.5 \,nm$.Then, $\frac{K_2}{K_1} = \left( \frac{\lambda_1}{\lambda_2} \right)^2 = \left( \frac{1}{0.5} \right)^2 = 2^2 = 4$.So, $K_2 = 4K_1$.The energy that should be added to the electron is $\Delta K = K_2 - K_1 = 4K_1 - K_1 = 3K_1$.Therefore, the energy to be added is three-times the initial energy.

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

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