JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 20)
An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio
of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c
= speed of light in vaccuum)
$${1 \over c}{\left( {{{2E} \over m}} \right)^{{1 \over 2}}}$$
$${1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$$
$${\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$$
$$c{\left( {2mE} \right)^{{1 \over 2}}}$$
Explanation
$$\lambda $$e = $${h \over {{p_e}}}$$ = $${h \over {\sqrt {2mE} }}$$
E = $${{hc} \over {{\lambda _{photon}}}}$$
$$ \Rightarrow $$ $${{\lambda _{photon}}}$$ = $${{hc} \over E}$$
$$ \therefore $$ $${{{\lambda _e}} \over {{\lambda _{photon}}}}$$ = $${1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$$
E = $${{hc} \over {{\lambda _{photon}}}}$$
$$ \Rightarrow $$ $${{\lambda _{photon}}}$$ = $${{hc} \over E}$$
$$ \therefore $$ $${{{\lambda _e}} \over {{\lambda _{photon}}}}$$ = $${1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$$
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