Question
An electron of mass $$m$$ and a photon have same energy $$E.$$ The ratio of de-Broglie wavelengths associated with them is
($$c$$ being velocity of light)
A.
$${\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$$
B.
$$c{\left( {2mE} \right)^{\frac{1}{2}}}$$
C.
$$\frac{1}{c}{\left( {\frac{{2m}}{E}} \right)^{\frac{1}{2}}}$$
D.
$$\frac{1}{c}{\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$$
Answer :
$$\frac{1}{c}{\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$$
Solution :
Since, it is given that electron has mass $$m.$$ de-Broglie’s wavelength for an electron will be given as
$${\lambda _e} = \frac{h}{P}\,......\left( {\text{i}} \right)$$
where, $$h$$ = Planck’s constant
$$P$$ = Linear momentum of electron
As kinetic energy of electron
$$E = \frac{{{P^2}}}{{2m}} \Rightarrow P = \sqrt {2mE} \,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$${\lambda _e} = \frac{h}{{\sqrt {2mE} }}\,......\left( {{\text{iii}}} \right)$$
Energy of a photon can be given as
$$\eqalign{
& E = hv \cr
& \Rightarrow E = \frac{{hc}}{{{\lambda _p}}} \cr
& \Rightarrow {\lambda _p} = \frac{{hc}}{E}\,......\left( {{\text{iv}}} \right) \cr} $$
Hence, $${\lambda _p} = $$ de-Broglie's wavelength of photon. Now, divide Eq. (iii) by Eq. (iv), we get
$$\eqalign{
& \frac{{{\lambda _e}}}{{{\lambda _p}}} = \frac{h}{{\sqrt {2mE} }} \cdot \frac{E}{{hc}} \cr
& \Rightarrow \frac{{{\lambda _c}}}{{{\lambda _p}}} = \frac{1}{c} \cdot \sqrt {\frac{E}{{2m}}} \cr} $$