Question

A proton, a neutron, an electron and an a-particle have same energy. Then their de Broglie wavelengths compare as

• A. ${\lambda }_p={\lambda }_n>{\lambda }_e>{\lambda }_{\alpha }$
• B. ${\lambda }_{\alpha }<{\lambda }_p={\lambda }_n<{\lambda }_e$
• C. ${\lambda }_e<{\lambda }_p={\lambda }_n>{\lambda }_{\alpha }$
• D. ${\lambda }_e={\lambda }_p={\lambda }_n={\lambda }_{\alpha }$

Answer 1

Answer: (B)

Kinetic energy of particle, $K=\frac{1}{2}m{v}^2$
or $mv=\sqrt{2mK}$
de Broglie wavelength,
$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mK}}$
For the given value of $K,\lambda \propto \frac{1}{\sqrt{m}}$
$\therefore {\lambda }_p:{\lambda }_n:{\lambda }_e:{\lambda }_{\alpha }=\frac{1}{\sqrt{m_p}}:\frac{1}{\sqrt{m_n}}:\frac{1}{\sqrt{m_e}}:\frac{1}{\sqrt{m_{\alpha }}}$
Since $m_p=m_n$, hence ${\lambda }_p={\lambda }_n$
As $m_{\alpha }>m_p$, therefore ${\lambda }_{\alpha }<{\lambda }_p$
As $m_e<m_n$, therefore ${\lambda }_e>{\lambda }_n$
Hence ${\lambda }_{\alpha }<{\lambda }_p={\lambda }_n<{\lambda }_e$