Question

In a photocell bichromatic light of wavelength $2475\,\mathring{A}$ and $6000 \,\mathring{A}$ 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 describe by the photoelectron will be (mass of electron $\left.=9 \times 10^{-31} kg \right)$

• A. 1 cm
• B. 5 cm
• C. 10 cm
• D. 25 cm

Answer 1

Answer: (B)

Energy of photons corresponding to light of wavelength
$\lambda_{1}=2475\,\mathring{A}$ is $E _{1}=\frac{12375}{2475}=5 eV$
and that corresponding to $\lambda_{2}=6000 \,\mathring{A}$ is
$E _{2}=\frac{12375}{6000}=2.06 eV$
As $E_{2}^{\prime} < W_{0}$ and $E_{1}>W_{0}$
Photoelectric emission is possible with $\lambda_{1}$ only. Maximum kinetic energy of emitted photoelectrons
$K=E-W_{0}=5-4.8=0.2 eV$
Photoelectrons experience magnetic force and move along a circular path of radius
$r =\frac{\sqrt{2 mk }}{ QB }=\frac{\sqrt{2 \times 9 \times 10^{-31} \times 0.2 \times 1.6 \times 10^{-19}}}{1.6 \times 10^{-19} \times 3 \times 10^{-5}}$
$=0.05 m =5$ cm