Be+3 and a proton are accelerated by the same potential their de Broglie wavelengths are in ratio of??

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Please find the solution to the asked query:
$$\begin{gathered} Charge of proton =e \\ mass of proton =u \\ Charge of B{e}^{3+} =3e \\ mass of Be particle =4u \\ De broglie wavelength is given by: \\ \lambda =\frac{h}{mv} \\ the momentum acquired by proton =\sqrt{2ueV} \\ thus {\lambda }_{proton} =\frac{h}{\sqrt{2ueV}} \\ for B{e}^{3+} particle: \\ \lambda { }_{B{e}^{3+}} =\frac{h}{\sqrt{2\times 4u\times 3e\times V}} \\ Then , \\ \frac{{\lambda }_{proton} }{\lambda { }_{B{e}^{3+}}}=\frac{\frac{h}{\sqrt{2ueV}}}{\frac{h}{\sqrt{2\times 4u\times 3e\times V}}}=\frac{1}{2\sqrt{3}} \end{gathered}$$

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