Question

Photoelectrons emitted from a photo sensitive metal of work function $1eV$ describe a circle of radius 0.1 cm in a magnetic field of induction ${10}^{-3}$ Tesla. The energy of the incident photons is (mass of electron $=9\times {10}^{-31}$kg)

• A. $1.17eV$
• B. $2.9eV$
• C. $0.9eV$
• D. $0.81eV$

Answer 1

The correct option is A $1.17eV$

$=\frac{mv}{r}=Bq$
$V=\frac{Bqr}{m}=Bq$
Velocity of electrons rotating i the magnetic field $=\frac{Bqn}{m}$
$=\frac{{10}^{-3}\times 1.6\times {10}^{-19}\times 0.1\times {10}^{-2}}{9.1\times {10}^{-31}}$ m/s
So,
energy of electron $=\frac{1}{2}m{v}^2=\frac{1}{2}\times \frac{B^2{e}^2{n}^2}{m^2}$
$=\frac{{10}^{-6}\times (1.6\times {10}^{-19})\times {10}^{-6}}{9.1\times {10}^{-31}}$ eV
$=\frac{1.6}{9}eV=0.17eV$
So,
$=$ energy of incident photon $=(1+0.17)eV$
$=1.17eV$