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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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four times initial energy

equal to the initial energy

thrice the initial energy

twice the initial energy

answer is B.

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Detailed Solution

$λ = \frac{h}{\sqrt{2 m K}} i.e. λ \propto \frac{1}{\sqrt{K}}$

$∴ \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{K_{2}}{K_{1}}} \Rightarrow \frac{1 \times 10^{- 9}}{0.5 \times 10^{- 9}} = \sqrt{\frac{K_{2}}{K_{1}}}$

$\Rightarrow K_{2} = 4 K_{1}$

$∴ Addition of energy = K_{2} - K_{1}$

$= 4 K_{1} - K_{1} = 3 K_{1}$

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