• question_answer Direction: Light having photons energy $hv$ is incident on a metallic plate having work function $\phi$ to eject the electrons. The most energetic electrons are then allowed to enter in a region of uniform magnetic field B as shown in the figure. The electrons are projected in XZ-plane making an angle of$\theta$ with X-axis and magnetic field is $\mathbf{B}={{B}_{0}}\,i$ along X-axis. Maximum pitch of the helix described by electron is found to be p. Take mass of electron as m and charge as q. Based on above information, answer the following questions The correct relation between P and ${{B}_{0}}$ is A)  $qP{{B}_{0}}=2\pi \,\cos \,\theta \,\sqrt{2\,(hv-\phi )\,m}$ B)  $qP{{B}_{0}}=2\pi \,\cos \,\theta \,\sqrt{\frac{2\,(hv-\phi )}{m}}$ C)  $qP{{B}_{0}}=2\pi \,\sqrt{2\,(hv-\phi )\,m}\,$ D)  $P=\frac{2gm}{q{{B}_{0}}}\times \sqrt{hv-\phi }$

At any time rthe location of electron is shown as P. In two dimensional view of electron in YZ-plane, the situation is more clear. As we kow that for photoelectron, $\frac{1}{2}\,m{{v}^{2}}=hv-\phi$ $\Rightarrow$            $v=\sqrt{\frac{2(hv-\phi )}{m}}$ And time period of change in magnetic field, $T=\frac{2\pi m}{qB}$ $\therefore$ Pitch, P = advance in on time period $v=\,\,\cos \,\,\theta \times t$ $\sqrt{\frac{2(hv-\phi )}{m}}\times \,\cos \,\theta \,\times \frac{2\pi m}{qB}$ $\Rightarrow$            $Pq{{B}_{0}}=2\pi \,\cos \,\theta \,\sqrt{2m(hv-\phi )}$
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