JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 16)
An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?
$${{{E_e}} \over {{E_{ph}}}} = {{2c} \over v}$$
$${{{E_e}} \over {{E_{ph}}}} = {v \over {2c}}$$
$${{{p_e}} \over {{p_{ph}}}} = {{2c} \over v}$$
$${{{p_e}} \over {{p_{ph}}}} = {v \over {2c}}$$
Explanation
$$\lambda = {h \over p} \Rightarrow p = {h \over \lambda }$$
Now, A/Q, $${h \over {{P_e}}} = {h \over {{P_{photon}}}}$$
$$ \Rightarrow {P_e} = {P_{photon}}$$ ....... (i)
Now, $${K_e} = {1 \over 2}M{v^2} = {{Pv} \over 2}$$
$${K_{ph}} = m{c^2} = Pc$$ ..... (ii)
$${{{K_e}} \over {{K_{eq}}}} = {v \over {2c}}$$
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