The de-Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is :

Let $\lambda$_p, $\lambda$_$\alpha$, m_p, m_$\alpha$, v_p, v_$\alpha$, p_p and p_$\alpha$ be the wavelength, mass, velocity and momentum of proton and $\alpha$-particle, respectively.<br/><br/>Given, $\lambda$_p = $\lambda$_$\alpha$<br/><br/>As we know that,<br/><br/>$\lambda$ = h/p<br/><br/>$\therefore$ ${h \over {{p_p}}} = {h \over {{p_\alpha }}}$<br/><br/>$\Rightarrow$ p_p = p_$\alpha$<br/><br/>$\Rightarrow$ m_pv_p = m_$\alpha$v_$\alpha$<br/><br/>$\Rightarrow$ m_pv_p = 4m_pv_$\alpha$ ($\because$ m_$\alpha$ = 4m_p)<br/><br/>$\Rightarrow$ ${{{v_p}} \over {{v_\alpha }}} = {4 \over 1}$ or 4 : 1