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 __________.

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: \(4\)