Question

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

(Assume mass of proton = mass of neutron)

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

Answer 1

Answer: (A)

Explanation: The de-Broglie wavelength for a mass m having a charge q when accelerated by a potential difference V is given by,

${\lambda }_b=\frac{h}{\sqrt{2mqV}}$

For the same potential,

${\lambda }_b\propto \frac{1}{\sqrt{qm}}$

For proton, let $q_1=q_p=e$ and $m_1=m_p=m$

Hence, $q_2=q_{{Be}^{3+}}=3e$ and $m_2=m_{{Be}^{3+}}=9m$

Therefore,

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{\sqrt{me}}{\sqrt{27me}}=\frac{1}{3\sqrt{3}}$

Hence, option (D) is correct.