Question

A proton, a neutron, an electron and an α-particle have same energy. Then, their de-Broglie wavelengths compare as(a) $\lambda_{p}=\lambda_{n}\ \textgreater \ \lambda_{e}\ \textgreater \ \lambda_{\alpha}$(b) $\lambda_{\alpha}\ \textless \ \lambda_{p}=\lambda_{n}\ \textgreater \ \lambda_{e}$(c) $\lambda_{e}\ \textless \ \lambda_{p}=\lambda_{n}\ \textgreater \ \lambda_{\alpha}$(d) $\lambda_{e}=\lambda_{p}=\lambda_{n}=\lambda_{\alpha}$

Answer 1

[B] is the correct option

Answer 2

Answer:

Explanation:

The de-Broglie Wavelength is inversely proportional to √ of mass and the kinetic energy. The kinetic energy being the same for all, thus the only thing left to compare is the √mass.

Energy of proton, neutron and electron = constant

K=1/2mv²

mv=√2ml

De- Broglie wave length is given by -

λ=h/√2mk

When k is constant =  λα/√1m

Thus, mp = mn

= λp = λn

= mα > mp = λα < λp

= me < mn = λe < λe

Thus, the de-Broglie wavelengths is compare as me < mn = λe < λe