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 $\mathrm{V}_2$ volt, the associated de Broglie wavelength is increased by $50 \%$. If $\left(V_1 / V_2\right)=(9 / \alpha)$, then the value of $\alpha$ is $\_\_\_\_$ .
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