A proton and $\displaystyle \alpha$-particle are accelerated through different potentials $\displaystyle \mathrm{V}_{1}$ and $\displaystyle \mathrm{V}_{2}$ respectively so that they have the same de Broglie wavelengths. Find $\displaystyle \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$.

De Broglie wavelength of a proton: $\displaystyle \lambda_{p}=\frac{h}{\sqrt{2 m_{p} q_{p} V_{1}}}=\frac{h}{\sqrt{2 m e V_{1}}}$ De Broglie wavelength of an $\displaystyle \alpha$ particle: $\displaystyle \lambda_{\alpha}=\frac{h}{\sqrt{2 m_{\alpha} q_{\alpha} V_{2}}}=\frac{h}{\sqrt{2(4 m)(2 e) V_{2}}}$ $\displaystyle \lambda_{p}=\lambda_{\alpha}$ $\displaystyle \frac{h}{\sqrt{2 m e V_{1}}}=\frac{h}{\sqrt{16 m e V_{2}}}$ $\displaystyle \frac{V_{1}}{V_{2}}=8$