The de Broglie wavelength for an electron accelerated through the...

The de Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \dfrac{h}{\sqrt{2meV}}$.

This implies $\lambda \propto \dfrac{1}{\sqrt{V}}$.

Given that the new wavelength $\lambda_2$ is increased by $50\%$, we have $\lambda_2 = \lambda_1 + 0.5\lambda_1 = \dfrac{3}{2}\lambda_1$.

Taking the ratio of the wavelengths, we get $\dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{V_2}{V_1}}$.

Substituting $\lambda_2 = \dfrac{3}{2}\lambda_1$, we get $\dfrac{2}{3} = \sqrt{\dfrac{V_2}{V_1}}$.

Squaring both sides yields $\dfrac{V_2}{V_1} = \dfrac{4}{9}$, which gives $\dfrac{V_1}{V_2} = \dfrac{9}{4}$.

Comparing this with the given relation $\dfrac{V_1}{V_2} = \dfrac{9}{\alpha}$, we find $\alpha = 4$.

Answer: $4$