JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 19)
An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)
$${1 \over c}{\left( {{{2m} \over E}} \right)^{1/2}}$$
$${1 \over c}{\left( {{E \over {2m}}} \right)^{1/2}}$$
$${\left( {{E \over {2m}}} \right)^{1/2}}$$
$$c{(2mE)^{1/2}}$$
Explanation
For photon, E = $${{hc} \over \lambda }$$
$${\lambda _p} = {{hc} \over E}$$ .... (i)
For electron, $${\lambda _e} = {{hc} \over {\sqrt {2mE} }}$$ .... (ii)
$${{{\lambda _e}} \over {{\lambda _p}}} = {{{{hc} \over {\sqrt {2mE} }}} \over {{{hc} \over E}}} = \sqrt {{E \over {2m{c^2}}}} = {1 \over c}{\left( {{E \over {2m}}} \right)^{1/2}}$$
$${\lambda _p} = {{hc} \over E}$$ .... (i)
For electron, $${\lambda _e} = {{hc} \over {\sqrt {2mE} }}$$ .... (ii)
$${{{\lambda _e}} \over {{\lambda _p}}} = {{{{hc} \over {\sqrt {2mE} }}} \over {{{hc} \over E}}} = \sqrt {{E \over {2m{c^2}}}} = {1 \over c}{\left( {{E \over {2m}}} \right)^{1/2}}$$
Comments (0)
