Dual Nature of matter and Radiation Part 2

The de Broglie wavelength $\lambda$ is given by:$\lambda = \frac{h}{p}$Where h is Planck's constant and p is the momentum.The kinetic energy ( $K$ ) of an electron accelerated through a potential $V$ is $K = e V = \frac{p^2}{2 m}$$p = \sqrt{2 m e V}$Substituting p into the wavelength formula:$\lambda = \frac{h}{\sqrt{2 m e V}}$For a given particle (electron), the wavelength is inversely proportional to the square root of the potential,$\lambda \propto \frac{1}{\sqrt{V}}$So, the ratio of wavelengths in two cases is,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$The initial wavelength is $\lambda_1$.The final wavelength is $\lambda_2 = \lambda_1 + 50\%$ of $\lambda_1 = 1.5 \lambda_1 = \frac{3}{2} \lambda_1$.Substituting these into the ratio:$\frac{\frac{3}{2} \lambda_1}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$$\Rightarrow \frac{3}{2} = \sqrt{\frac{V_1}{V_2}}$$\Rightarrow \left(\frac{3}{2}\right)^2 = \frac{V_1}{V_2}$$\Rightarrow \frac{V_1}{V_2} = \frac{9}{4} = \frac{9}{\alpha} \Rightarrow \alpha = 4$Therefore, the value of $\alpha$ is 4 .