Question

If an electron, a proton and an alpha particle have same de Broglie wavelengths then which statement about their kinetic energies is correct?

• A. Kinetic energy of alpha particle is least
• B. Kinetic energy of proton is least
• C. Kinetic energy of electron is least
• D. Kinetic energies of all three are same

Answer 1

Answer: (A)

de Broglie wavelength of a particle of mass $m$ and kinetic energy $K$ is
$\lambda =\frac{h}{\sqrt{2mk}}$
$\therefore {\lambda }_e=\frac{h}{\sqrt{2m_e{K}_e}}$, ${\lambda }_p=\frac{h}{\sqrt{2m_p{K}_p}}$
and ${\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }{K}_{\alpha }}}$
where subscripts $e,p$ and $\alpha$ represent electron, proton and alpha particle respectively But ${\lambda }_e={\lambda }_p={\lambda }_{\alpha }$ (given)
$\therefore \frac{h}{\sqrt{2m_e{K}_e}}=\frac{h}{\sqrt{2m_p{K}_p}}=\frac{h}{\sqrt{2m_{\alpha }{K}_{\alpha }}}$
or $m_e{K}_e=m_p{K}_p=m_{\alpha }{K}_{\alpha }$
As $m_e<m_p<m_{\alpha }$
$\therefore K_e>K_p>K_{\alpha }$
Thus, the kinetic energy of alpha particle is least