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

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$\frac{7}{9} E$

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(B) The de-Broglie wavelength $\lambda$ is related to kinetic energy $E$ by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.From this,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$.Let the initial wavelength be $\lambda_1 = \lambda$ and initial energy be $E_1 = E$. The final wavelength is $\lambda_2 = 0.75 \lambda_1 = \frac{3}{4} \lambda_1$.Using the proportionality $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{E_1}{E_2}}$,we get $\frac{3}{4} = \sqrt{\frac{E}{E_2}}$.Squaring both sides,we get $\frac{9}{16} = \frac{E}{E_2}$,which implies $E_2 = \frac{16}{9} E$.The extra energy required is $\Delta E = E_2 - E_1 = \frac{16}{9} E - E = \frac{7}{9} E$.

The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. 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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