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(D) The de Broglie wavelength $\lambda$ is related to the kinetic energy $E$ by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.From this, we can expres

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Three times its initial energy

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(D) The de Broglie wavelength $\lambda$ is related to the kinetic energy $E$ by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.From this, we can express energy as $E = \frac{h^2}{2m\lambda^2}$.Let the initial wavelength be $\lambda_1 = 1 \,nm$ and the final wavelength be $\lambda_2 = 0.5 \,nm$.The initial energy is $E_1 = \frac{h^2}{2m\lambda_1^2}$.The final energy is $E_2 = \frac{h^2}{2m\lambda_2^2} = \frac{h^2}{2m(0.5\lambda_1)^2} = \frac{h^2}{2m(0.25\lambda_1^2)} = 4E_1$.The additional energy required is $\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$.Thus, the additional energy to be supplied is three times its initial energy.

What is the additional energy that should be supplied to a moving electron to reduce its de Broglie wavelength from $1 \,nm$ to $0.5 \,nm$?

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