Question

If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal then the ratio of velocities of the proton and the $\alpha$-particle will be:

• A. $4:1$
• B. $2:1$
• C. $1:2$
• D. $1:4$

Answer 1

The correct option is A $4:1$

De - Broglie wavelength

$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$

wherein $-h=$ plank's constant

$m=$ mass of particle, $v=$ speed of particle

$E=$ Kinetic energy of particle

$\lambda =\frac{h}{mv}$ $\because$ ${\lambda }_{\alpha }={\lambda }_p$ $\therefore$ $p_a=p_b$

$\therefore$ $m\alpha v_{\alpha }=m_p{v}_p$

$\therefore$ $\frac{v_p}{v_{\alpha }}=\frac{m_a}{m_p}=4:1$