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The de-Broglie wavelength of a particle having kinetic energy E is $λ$ . How much extra energy must be given to this particle so that the de- Broglie wavelength reduces to 75% of the initial value.

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$\frac{7}{9} E$

$\frac{16}{9} E$

$\frac{1}{9} E$

answer is B.

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Detailed Solution

According to de-Broglie hypothes is $λ = \frac{h}{p} = \frac{h}{\sqrt{2 m E}} \Rightarrow λ \propto \frac{1}{\sqrt{E}}$

$\frac{λ_{2}}{λ_{1}} = \sqrt{\frac{E_{1}}{E_{2}}} = \frac{3}{4} \Rightarrow λ_{2} = 0.75 λ_{1}$ $\frac{E_{1}}{E_{2}} = \frac{16}{9} \Rightarrow E_{1} = \frac{16}{9} E$

So, extra energy $=$ $\frac{16}{9} E - E \Rightarrow \frac{7}{9} E$

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