Question

A proton, a neutron, an electron and an $\alpha$-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

The correct option is B ${\lambda }_{\alpha }<{\lambda }_p={\lambda }_n<{\lambda }_e$

Kinetic energy of particle, $K=\frac{1}{2}m{v}^{2}ormv=\sqrt{2mK}$

de Broglie wavelength, $\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mK}}$

For the given value of $K,\lambda \alpha \frac{1}{\sqrt{m}}$

$\therefore {\lambda }_p:{\lambda }_n:{\lambda }_e:{\lambda }_{\alpha }=\frac{1}{\sqrt{m_p}}:\frac{1}{\sqrt{m_n}}:\frac{1}{\sqrt{m_e}}:\frac{1}{\sqrt{m_{\alpha }}}$

Since $m_p=m_n,hence{\lambda }_p={\lambda }_n$

$As{m}_{\alpha }>m_p,therefore{\lambda }_{alpha}<{\lambda }_p$

As $m_e<m_n,therefore{\lambda }_n<{\lambda }_n$

Hence ${\lambda }_{alpha}<{\lambda }_p={\lambda }_n<{\lambda }_e$