MHT CET · Physics · Dual Nature of Matter
A photon of energy ' \(E\) ' ejects photoelectrons from a metal surface whose work function is \(W_0\). If this electron enters into a uniform magnetic field with induction ' \(B\) ' in a direction perpendicular to the field and describes a circular path of radius ' \(r\) ', then the radius is given by
- A \(\sqrt{\frac{2 e\left(E-W_0\right)}{m B}}\)
- B \(\frac{\sqrt{2\left(E-W_0\right) m}}{e B}\)
- C \(\sqrt{\frac{2 m\left(E-W_0\right)}{m B}}\)
- D \(\sqrt{2 m\left(E-W_0\right) e B}\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{2\left(E-W_0\right) m}}{e B}\)
Step-by-step Solution
Detailed explanation
Concept: Photoelectric effect and the motion of charge in a direction perpendicular to the magnetic field`
The maximum kinetic energy \(K=\frac{m v^2}{2}\) of a photo-electron is given by,'
\(K=E-W_0\)
where, \(E\) is the energy of incident photon and \(W_0\) is the work function of the metal.
Therefore, the velocity of the photoelectron is: \(v=\sqrt{\frac{2\left(E-W_0\right)}{m}}\)
The photoelectron undergoes uniform circular motion as its velocity is perpendicular to the magnetic field. On balancing magnetic force on moving charge with centrifugal force:
\(q v B=\frac{m v^2}{r}\)
Therefore, \(r=\frac{m v}{B q}=\frac{\sqrt{2\left(E-W_0\right) m}}{e B}\)
The maximum kinetic energy \(K=\frac{m v^2}{2}\) of a photo-electron is given by,'
\(K=E-W_0\)
where, \(E\) is the energy of incident photon and \(W_0\) is the work function of the metal.
Therefore, the velocity of the photoelectron is: \(v=\sqrt{\frac{2\left(E-W_0\right)}{m}}\)
The photoelectron undergoes uniform circular motion as its velocity is perpendicular to the magnetic field. On balancing magnetic force on moving charge with centrifugal force:
\(q v B=\frac{m v^2}{r}\)
Therefore, \(r=\frac{m v}{B q}=\frac{\sqrt{2\left(E-W_0\right) m}}{e B}\)
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