Question

The magnitude of the de-Broglie wavelength $\left(\lambda\right)$ of electron (e), proton (p), neutron (n) and a-particle (a) all having the same energy of 1 MeV, 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)

$\lambda_{\alpha}, \lambda_{n}. \lambda_{p}, \lambda_{e}$

Answer 2

$\lambda=\frac{h}{\sqrt{2mE}} So \lambda \propto\frac{1}{\sqrt{m}}$
since $m_{\alpha}>m_{n}>m_{p}>m_{e}$
so de-Broglie wave length in increasingorder will be $\lambda_{\alpha}