How many volts?

Note that an alpha particle has a charge of $+2$ and consists of two protons and two neutrons which have approximately the same mass. To solve this problem, we use the De Broglie equation, which states: $\lambda = \frac{h}{p}=\frac{h}{mv}$ . First, we well derive an equation for the velocity that a particle will have after passing through an electric current:

$\begin{aligned} E_{el}&=E_{kin} \\ Uq &= \frac{1}{2}mv^2 \\ v &= \sqrt{\frac{2Uq}{m}} \end{aligned}$

So, for the proton we have:

$\begin{aligned} \lambda_{p^+} = \frac{h}{m_ev} = \frac{h}{m_e\sqrt{\frac{2U_{p^+}e}{m_e}}}=\frac{h}{m_e\sqrt{2U_{p^+}em_e}} \end{aligned}$

And for the alpha particle:

$\begin{aligned} \lambda_{\alpha} = \frac{h}{4m_ev} = \frac{h}{4m_e\sqrt{\frac{2U_\alpha(2e)}{4m_e}}}= \frac{h}{4m_e\sqrt{U_\alpha e m_e}} \end{aligned}$

We need to set $\lambda_{p^+}=\lambda_{\alpha}$ . From this, we can derive the following:

$\begin{aligned} m_e\sqrt{2U_{p^+}em_e}&=4m_e\sqrt{U_\alpha e m_e} \\ 2U_{p^+}em_e&=16U_\alpha e m_e \\ U_{p^+} &= 8 U_\alpha \end{aligned}$

Thus, answer we are looking for is $\boxed{\frac{1}{8}U_{p^+}}$