A metal surface having work function ' $\mathrm{w}_{0}{ }^{\prime}$ emits photoelectrons when photons of…

A metal surface having work function ' $\mathrm{w}_{0}{ }^{\prime}$ emits photoelectrons when photons of energy 'E' are incident on it. The electron enters the uniform magnetic field (B) in perpendicular direction and moves in circular path of radius 'r'. Then 'r' is equal to (m and e be the mass and charge of electron respectively).
  1. $\frac{\sqrt{m\left(E-W_{0}\right)}}{e B}$
  2. $\frac{m\left(\mathrm{E}-\mathrm{W}_{\mathrm{O}}\right)}{\mathrm{eB}}$
  3. $\frac{\sqrt{2 m\left(E-W_{0}\right)}}{\mathbf{e B}}$
  4. $\frac{2 m\left(E-w_{0}\right)}{e B}$

Solution

Kinetic energy of the electron $\mathrm{K}=\mathrm{E}-\omega_{0}$ Momentum $P=\sqrt{2 m k}=\sqrt{2 m\left(E-\omega_{\mathrm{o}}\right)}$ Radius of the circular path $r=\frac{P}{e B}=\frac{\sqrt{2 m\left(E-\omega_{0}\right)}}{e B}$

Asked in: MHT CET 2020 (14 Oct Shift 1)

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