Medium MCQ +4 / -1 PYQ · JEE Mains 2022

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?

  1. A ${{{E_e}} \over {{E_{ph}}}} = {{2c} \over v}$
  2. B ${{{E_e}} \over {{E_{ph}}}} = {v \over {2c}}$ Correct answer
  3. C ${{{p_e}} \over {{p_{ph}}}} = {{2c} \over v}$
  4. D ${{{p_e}} \over {{p_{ph}}}} = {v \over {2c}}$

Solution

<p>$\lambda = {h \over p} \Rightarrow p = {h \over \lambda }$</p> <p>Now, A/Q, ${h \over {{P_e}}} = {h \over {{P_{photon}}}}$</p> <p>$\Rightarrow {P_e} = {P_{photon}}$ ....... (i)</p> <p>Now, ${K_e} = {1 \over 2}M{v^2} = {{Pv} \over 2}$</p> <p>${K_{ph}} = m{c^2} = Pc$ ..... (ii)</p> <p>${{{K_e}} \over {{K_{eq}}}} = {v \over {2c}}$</p>

About this question

Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis

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