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 ?

Given, initial kinetic energy, E₁ = E

Initial de-Broglie wavelength, λ₁ = λ

Consider the wavelength of particle changes to 75% of λ after providing energy ∆E to the particle.

Hence, final wavelength of particle, λ₂ = 0.75λ

Final energy, ${\mathrm{E}}_2=\mathrm{E}+∆\mathrm{E}$

Relationship between de-Broglie wavelength and energy of particle is given as

            $\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$

Here, h is Planck’s constant and m is mass of particle which is also constant term.

Therefore, we get

             $\mathrm{\lambda }\propto \frac{1}{\sqrt{\mathrm{E}}}$

Thus, we can write the relationship as

$\begin{array}{l} \frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2}=\frac{\sqrt{{\mathrm{E}}_2}}{\sqrt{{\mathrm{E}}_1}}\\ \Rightarrow \frac{\mathrm{\lambda }}{0.75\mathrm{\lambda }}=\frac{\sqrt{\mathrm{E}+\mathrm{\Delta E}}}{\sqrt{\mathrm{E}}}\\ \Rightarrow \frac{4}{3}=\frac{\sqrt{\mathrm{E}+\mathrm{\Delta E}}}{\sqrt{\mathrm{E}}}\end{array}$

Squaring both sides of above equation, we get

$\begin{array}{l} \frac{16}{9}=\frac{\mathrm{E}+\mathrm{\Delta E}}{\mathrm{E}}\\ \Rightarrow 16\mathrm{E}=9\mathrm{E}+9\mathrm{\Delta E}\\ \Rightarrow \mathrm{\Delta E}=\frac{7}{9}\mathrm{E}\end{array}$