JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
A monochromatic light is incident on a metallic plate having work function \(\phi\). An electron, emitted normally to the plate from a point \(A\) with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point \(B\). The distance between A and B is :
(Given : The magnitude of charge of an electron is e and mass is \(\mathrm{m}, \mathrm{h}\) is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)
- A \(\sqrt{2 \mathrm{~m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}\)
- B \(\sqrt{\mathrm{m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}\)
- C \(\sqrt{8 \mathrm{~m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}\)
- D \(2 \sqrt{\mathrm{~m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{8 \mathrm{~m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{KE}_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi \\ & \mathrm{p}=\sqrt{2 \mathrm{mK}_{\max }} \\ & \mathrm{p}=\sqrt{2 \mathrm{~m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} \\ & \mathrm{d}_{\mathrm{A}-\mathrm{B}}=2 \mathrm{R} \\ &…
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