Question

The magnitude of the de-Broglie wavelength $(\lambda )$ of electron $(e)$, proton $(p)$, neutron $(n)$ and $\alpha$-particle $(a)$ all having the same energy of $1MeV$, in the increasing order will follow the sequence

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

Answer 1

Answer: (C)

$\lambda =\frac{h}{\sqrt{2mE}}So\lambda \propto \frac{1}{\sqrt{m}}$
since $m_{\alpha }>m_n>m_p>m_e$
so de-Broglie wave length in increasingorder will be ${\lambda }_{\alpha }<{\lambda }_n<{\lambda }_e$