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Relation between wavelength of photon and electron of same energy is

Question :- Relation between wavelength of photon and electron of same energy is

(A) $\displaystyle {{\lambda }_{ph}}>{{\lambda }_{e}}$

(B) $\displaystyle {{\lambda }_{ph}}<{{\lambda }_{e}}$

(D) $\displaystyle \frac{{{\lambda }_{e}}}{{{\lambda }_{ph}}}=\text{constant}$

Solution :-

As de Broglie wavelength associated with an electron is given by

$\displaystyle {{\lambda }_{e}}=\frac{h}{\sqrt{2{{m}_{e}}E}}$ …..(1)

And for photon

$\displaystyle E=h\nu =\frac{hc}{{{\lambda }_{ph}}}$ …..(2)

Using this value in equation (1) we get,

$\displaystyle {{\lambda }_{e}}=\frac{h}{\sqrt{2{{m}_{e}}\frac{hc}{{{\lambda }_{ph}}}}}=\sqrt{\frac{h{{\lambda }_{ph}}}{2{{m}_{e}}c}}$

$\displaystyle \Rightarrow {{\lambda }_{ph}}=\frac{2{{m}_{e}}c}{h}{{\left( {{\lambda }_{e}} \right)}^{2}}$

Now order of $\displaystyle \frac{2{{m}_{e}}c}{h}$ is about 10¹¹, hence $\displaystyle {{\lambda }_{ph}}>{{\lambda }_{e}}$ .

So option (A) is correct.