The de Broglie wavelength for an electron accelerated through the potential difference of $$V_1$$ volt is $$\lambda_1$$. When the potential difference is changed to $$V_2$$ volt, the associated de Broglie wavelength is increased by 50%. If $$(V_1 / V_2) = (9 / \alpha)$$, then the value of $$\alpha$$ is __________.

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_e eV}}$$

$$\lambda \propto \frac{1}{\sqrt{V}}$$

$$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$$

$$\left(\frac{\lambda_1}{\lambda_2}\right)^2 = \frac{V_2}{V_1} \implies \frac{V_1}{V_2} = \left(\frac{\lambda_2}{\lambda_1}\right)^2 \quad \text{--- (Equation 1)}$$

$$\lambda_2 = \lambda_1 + 50\% \text{ of } \lambda_1$$

$$\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1 = \frac{3}{2}\lambda_1$$

$$\frac{\lambda_2}{\lambda_1} = \frac{3}{2}$$

$$\frac{V_1}{V_2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$

$$\frac{V_1}{V_2} = \frac{9}{\alpha}$$

$$\frac{9}{4} = \frac{9}{\alpha} \implies \alpha = 4$$