Question

Calculate the ratio of wavelength for an $\alpha$ particle and proton accelerated through same potential difference.

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

Answer 1

Answer: (C)

Using de Broglie equation $\lambda =\frac{h}{\sqrt{2qmV}}$

$\lambda =$ wavelength, q = charge, m = mass, V = Potential and h = Plank constant

${\lambda }_a=\frac{h}{\sqrt{2q_a{m}_{a}V}}$ ${\lambda }_p=\frac{h}{\sqrt{2q_p{m}_{p}V}}$

$m_a=4m_p,q_a=2q_p$

$\frac{{\lambda }_a}{{\lambda }_p}=\sqrt{\frac{q_p{m}_p}{q_a{m}_a}}=\sqrt{\frac{1}{2}.\frac{1}{4}}=\sqrt{\frac{1}{8}}$

$\frac{{\lambda }_a}{{\lambda }_p}=\frac{1}{2\sqrt{2}}$