Question

A proton and an $\alpha$-particle are accelerated through the same potential difference. Which one of the two has (i) greater de-Broglie wavelength, and (ii) less kinetic energy? Justify your answer.

Answer 1

(i) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $e$ , accelerated by a potential difference $V$ is given by ,

$\lambda =\frac{h}{\sqrt{2meV}}$

for proton , ${\lambda }_p=\frac{h}{\sqrt{2m_p{e}_{p}V}}$ ....................eq1

for alpha , ${\lambda }_a=\frac{h}{\sqrt{2m_a{e}_{a}V}}$ .....................eq2

dividing eq1 by eq2 we get ,

$\frac{{\lambda }_p}{{\lambda }_a}=\sqrt{\frac{m_a{e}_a}{m_p{e}_p}}$

now we know that , $m_a=4m_p,e_a=2e_p$

therefore $\frac{{\lambda }_p}{{\lambda }_a}=\sqrt{\frac{4m_p\times 2e_p}{m_p{e}_p}}=\sqrt{8}$

or ${\lambda }_p>{\lambda }_a$

(ii) de-Broglie wavelength in terms of kinetic energy is given by ,

$\lambda =\frac{h}{\sqrt{2mK}}$ ,

or $K=h^2/2m{\lambda }^2$

for proton $K_p=h/2m_p{\lambda }_p^2$

for alpha $K_a=h/2m_a{\lambda }_a^2$

or $\frac{K_p}{K_a}=\frac{m_{a{\lambda }_a^2}}{m_p{\lambda }_p^2}$

or $\frac{K_p}{K_a}=\frac{4m_p{\lambda }_a^2}{m_p\times 8{\lambda }_a^2}=1/2$

or $K_p<K_a$