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

Please find the solution to the asked query:

$Chargeofproton=e\phantom{\rule{0ex}{0ex}}massofproton=u\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}ChargeofB{e}^{3+}=3e\phantom{\rule{0ex}{0ex}}massofBeparticle=4u\phantom{\rule{0ex}{0ex}}Debrogliewavelengthisgivenby:\phantom{\rule{0ex}{0ex}}\lambda =\frac{h}{mv}\phantom{\rule{0ex}{0ex}}themomentumacquiredbyproton=\sqrt{2ueV}\phantom{\rule{0ex}{0ex}}thus{\lambda}_{proton}=\frac{h}{\sqrt{2ueV}}\phantom{\rule{0ex}{0ex}}forB{e}^{3+}particle:\phantom{\rule{0ex}{0ex}}\lambda {}_{B{e}^{3+}}=\frac{h}{\sqrt{2\times 4u\times 3e\times V}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Then,\phantom{\rule{0ex}{0ex}}\frac{{\lambda}_{proton}}{\lambda {}_{B{e}^{3+}}}=\frac{\frac{h}{\sqrt{2ueV}}}{\frac{h}{\sqrt{2\times 4u\times 3e\times V}}}=\frac{1}{2\sqrt{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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