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(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m E_{k}}}$.Since $h$ and $m$ are constants,we have $\lambda \propto \frac{1}{\sq

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$3$ times the initial kinetic energy

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(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m E_{k}}}$.Since $h$ and $m$ are constants,we have $\lambda \propto \frac{1}{\sqrt{E_{k}}}$.Let the initial kinetic energy be $E_{k1} = E$ and the final kinetic energy be $E_{k2}$.Given $\lambda_1 = 1 \ nm$ and $\lambda_2 = 0.5 \ nm$,we have $\frac{\lambda_1}{\lambda_2} = \frac{1}{0.5} = 2$.Using the relation $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{E_{k2}}{E_{k1}}}$,we get $2 = \sqrt{\frac{E_{k2}}{E}}$,which implies $\frac{E_{k2}}{E} = 4$,so $E_{k2} = 4E$.The additional energy required is $\Delta E = E_{k2} - E_{k1} = 4E - E = 3E$.Thus,the additional energy is $3$ times the initial kinetic 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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