Question

A photon of energy ‘E’ ejects photoelectrons from a metal surface whose work function is W₀. If this electron enters into uniform magnetic field of induction ‘B’ in a direction perpendicular to field and describes a circular path of radius‘r’, then radius is given by

• A. \(\frac {\sqrt {2m(E-W_0)}}{eB}\)
• B. \(\frac {2e(E-W_0)}{mB}\)
• C. \(\sqrt {\frac {2m(E-W_0)}{eB}}\)
• D. \(\sqrt {2m(E-W_0)eB}{}\)

Answer 1

The Correct Option is A

The equation for the radius of a charged particle moving in a magnetic field is given by: 
r = \(\frac {mv}{eB}\), 
To find the velocity of the electron, we can use the conservation of energy. The energy of the photon (E) is used to eject the photoelectron from the metal surface, so the remaining energy is given by E - W₀, where W₀ is the work function of the metal. 
Since the electron is not subject to any external forces (apart from the magnetic force), the remaining energy is converted into the kinetic energy of the electron: 
E - W₀ = \(\frac {1}{2}\)mv². 
Rearranging this equation, we have: 
v =\(\sqrt {\frac {2(E - W_0)}{m}}\). 
Substituting this expression for v into the equation for the radius, we get: 
r = \(\frac {m.\sqrt{2(E - W0)} / m}{eB}\). 
Simplifying, we find: 
r = \(\frac {\sqrt {2m(E-W_0)}}{eB}\). 
Therefore, the correct option is (A) \(\frac {\sqrt {2m(E-W_0)}}{eB}\)