JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 13)
An electron moving with speed v and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is :
$${{3c} \over v}$$
$${v \over {3c}}$$
$${v \over {2c}}$$
$${{2c} \over v}$$
Explanation
$${\lambda _e} = {\lambda _{Ph}}$$
$${h \over {{p_e}}} = {h \over {{p_{ph}}}}$$
$$\sqrt {2m{k_e}} = {{{E_{ph}}} \over c}$$
$$2m{k_e} = {{{{({E_{ph}})}^2}} \over {{c^2}}}$$
$${{{k_e}} \over {{E_{ph}}}} = {{{E_{ph}}} \over {{c^2}}}\left( {{1 \over {2m}}} \right)$$
$$ = {{{p_{ph}}} \over c}\left( {{1 \over {2m}}} \right)$$
$$ = {{{p_e}} \over c}\left( {{1 \over {2m}}} \right)$$
$$ = {{mv} \over c}{1 \over {2m}}$$
$$ = {v \over {2c}}$$
$${h \over {{p_e}}} = {h \over {{p_{ph}}}}$$
$$\sqrt {2m{k_e}} = {{{E_{ph}}} \over c}$$
$$2m{k_e} = {{{{({E_{ph}})}^2}} \over {{c^2}}}$$
$${{{k_e}} \over {{E_{ph}}}} = {{{E_{ph}}} \over {{c^2}}}\left( {{1 \over {2m}}} \right)$$
$$ = {{{p_{ph}}} \over c}\left( {{1 \over {2m}}} \right)$$
$$ = {{{p_e}} \over c}\left( {{1 \over {2m}}} \right)$$
$$ = {{mv} \over c}{1 \over {2m}}$$
$$ = {v \over {2c}}$$
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