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