Question

${Be}^{3+}$ and a proton are accelerated by same proton, their de-broglie wavelengths are in the ratio:

(Assume the mass of proton=mass of neutron)

• A. 1:1
• B. 1:2
• C. 1:4
• D. 1:$3\sqrt{3}$

Answer 1

The correct option is D 1:$3\sqrt{3}$

${\lambda }_p=\frac{h}{\sqrt{2eV{m}_p}}$

${\lambda }_{B{e}^{3+}}=\frac{h}{\sqrt{3\times 3eV{m}_{B{e}^{3+}}}}$

$=\frac{h}{\sqrt{2\times 3eV\times 9m_p}}$

Hence,

$\frac{{\lambda }_{B{e}^{3+}}}{{\lambda }_p}=\sqrt{\frac{2eV{m}_p}{2\times 3eV\times 9m_p}}$

$=\frac{1}{3\sqrt{3}}$