Dual Nature of Matter and Radiation Part 1

Given, initial kinetic energy, $E_{1}=E$ Initial de-Broglie wavelength, $\lambda_{1}=\lambda$ Consider the wavelength of particle changes to $75 \%$ of λ after providing energy ∆E to the particle. Hence, final wavelength of particle, $\lambda_{2}=0.75 \lambda$ Final energy, $E_{2}=E+\Delta E$ Relationship between de-Broglie wavelength and energy of particleis given as $\lambda=\frac{h}{\sqrt{2 m E}}$ Here, h is Planck’s constant and m is mass of particle which is alsoconstant term. Therefore, we get $\lambda \propto \frac{1}{\sqrt{E}}$ Thus, we can write the relationship as $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\sqrt{E_{2}}}{\sqrt{E_{1}}}$ $\Rightarrow \frac{\lambda}{0.75 \lambda}=\frac{\sqrt{E+\Delta E}}{\sqrt{E}}$ $\Rightarrow \frac{4}{3}=\frac{\sqrt{E+\Delta E}}{\sqrt{E}}$ Squaring both sides of above equation, we get $\Rightarrow \frac{16}{9}=\frac{E+\Delta E}{E}$ $\Rightarrow 16E=9E+9\Delta E$ $\Delta E=\frac{7}{9}E$