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
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mk}}$
$\lambda \propto \frac{1}{\sqrt{m}}$ (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, ${\lambda }_e>{\lambda }_n>{\lambda }_{\alpha }$