Q. In a photocell bichromatic light of wavelength $2475\ \mathring{\mathrm{A}}$ and $6000\ \mathring{\mathrm{A}}$ are incident on cathode whose work function is $4.8\ eV.$ If a uniform magnetic field of $3 \times 10^{- 5}tesla$ exists parallel to the plate, the radius of the path described by the photoelectron will be (mass of electron $= 9 \times 10^{- 31}kg$)
Explanation
(b) Energy of photons corresponding to light of wavelength $\lambda_{1} = 2475\ \mathring{\mathrm{A}}$ is$\ E_{1} = \frac{12375}{2475} = 5\ eV$
and that corresponding to $\lambda_{2} = 6000\ \mathring{\mathrm{A}}\ is$
$E_{2} = \frac{12375}{6000} = 2.06\ eV$
As $E_{2} < W_{0}$ and $E_{1} > W_{0}.$ Photoelectric emission is possible with $\lambda_{1}$ only. Maximum kinetic energy of emitted photoelectrons $K = E - W_{0} = 5 - 4.8 = 0.2\ eV.$
Photo electrons experience magnetic force and move along a circular path of radius
$r = \frac{\sqrt{2mk}}{Q\ B} = \frac{\sqrt{2 \times 9 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}{1.6 \times 10^{- 19} \times 3 \times 10^{- 5}}$
$= 0.05\ m = 5\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.