A metal surface is illuminated by a radiation of wavelength $$4500$$ Å. The ejected photo-electron enters a constant magnetic field of $$2$$ mT making an angle of $$90°$$ with the magnetic field. If it starts revolving in a circular path of radius $$2$$ mm, the work function of the metal is approximately

We need to find the work function of the metal using the photoelectric effect and circular motion in a magnetic field.

Wavelength of radiation: $$\lambda = 4500$$ Å $$= 4500 \times 10^{-10}$$ m

Magnetic field: $$B = 2$$ mT $$= 2 \times 10^{-3}$$ T

Radius of circular path: $$r = 2$$ mm $$= 2 \times 10^{-3}$$ m

Angle with magnetic field = 90°

For a charged particle moving in a circle in a magnetic field:

$$r = \frac{mv}{eB}$$

$$mv = eBr$$

$$mv = 1.6 \times 10^{-19} \times 2 \times 10^{-3} \times 2 \times 10^{-3}$$

$$mv = 6.4 \times 10^{-25} \text{ kg m s}^{-1}$$

$$KE = \frac{(mv)^2}{2m} = \frac{(6.4 \times 10^{-25})^2}{2 \times 9.1 \times 10^{-31}}$$

$$KE = \frac{40.96 \times 10^{-50}}{18.2 \times 10^{-31}} = \frac{40.96}{18.2} \times 10^{-19}$$

$$KE = 2.25 \times 10^{-19} \text{ J}$$

$$KE = \frac{2.25 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.406 \text{ eV}$$

$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4500 \times 10^{-10}}$$

$$E = \frac{19.89 \times 10^{-26}}{4.5 \times 10^{-7}} = 4.42 \times 10^{-19} \text{ J}$$

$$E = \frac{4.42 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.76 \text{ eV}$$

$$E = \phi + KE$$

$$\phi = E - KE = 2.76 - 1.406 \approx 1.36 \text{ eV}$$

Hence, the correct answer is Option A.