Question

An electron, a neutron and an alpha particle have same kinetic energy and their de-Broglie wavelength are $\lambda _{e}, \, \lambda _{n}$ and $\lambda _{\alpha }$ respectively. Which statement is correct about their de-Broglie wavelengths?

• A. $\lambda _{e}>\lambda _{n}>\lambda _{\alpha }$
• B. $\lambda _{e} < \lambda _{n}>\lambda _{\alpha }$
• C. $\lambda _{e} < \lambda _{n} < \lambda _{\alpha }$
• D. $\lambda _{e}>\lambda _{n} < \lambda _{\alpha }$

Answer 1

Answer: (A)

According to de $-$ Broglie theory, the wavelength
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2 m k}}$
$\lambda \propto \frac{1}{\sqrt{m}}$ (for same kinetic energy)
The wavelength is inversely proportional to the root of mass of the particle.
As $m_{e} < m_{n} < m_{\alpha }$
So, $\lambda _{e}>\lambda _{n}>\lambda _{\alpha }$