A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity $$B$$. The electrons emitted perpendicular to the magnetic field $$B$$, with maximum kinetic energy are bent into a circular arc of radius 30 cm. The value of $$B$$ is _________ $$\times 10^{-7}$$ T.
Given $$hc = 20 \times 10^{-26}$$ J m, the mass of the electron = $$9 \times 10^{-31}$$ kg.

We have a photon of wavelength $$\lambda = 500 \text{ nm}=500 \times 10^{-9}\ \text{m}.$$ The energy of one photon is obtained from the formula $$E=\dfrac{hc}{\lambda}.$$ Given $$hc = 20 \times 10^{-26}\ \text{J m},$$ we substitute:

$$E=\dfrac{20 \times 10^{-26}}{500 \times 10^{-9}} =\dfrac{20}{5}\times 10^{-26+7} =4 \times 10^{-19}\ \text{J}.$$

The work function of the metal is $$\phi =1.25\ \text{eV}.$$ Since $$1\ \text{eV}=1.6 \times 10^{-19}\ \text{J},$$

$$\phi =1.25 \times 1.6 \times 10^{-19} =2.0 \times 10^{-19}\ \text{J}.$$

Einstein’s photo-electric equation states $$\text{K.E.}_{\max}=E-\phi.$$ So,

$$\text{K.E.}_{\max}=4 \times 10^{-19}-2 \times 10^{-19} =2 \times 10^{-19}\ \text{J}.$$

Kinetic energy and speed are related by $$\text{K.E.}=\dfrac{1}{2}mv^{2}.$$ Therefore

$$v=\sqrt{\dfrac{2\,\text{K.E.}_{\max}}{m}} =\sqrt{\dfrac{2 \times 2 \times 10^{-19}} {9 \times 10^{-31}}} =\sqrt{\dfrac{4}{9}\times 10^{12}} =\sqrt{0.444\;4 \times 10^{12}} \approx 6.66 \times 10^{5}\ \text{m s}^{-1}.$$

An electron moving perpendicular to a magnetic field describes a circle of radius $$r$$ given by $$r=\dfrac{mv}{eB},$$ where $$e=1.6 \times 10^{-19}\ \text{C}.$$ Rearranging,

$$B=\dfrac{mv}{er}.$$

The radius is $$r = 30 \text{ cm}=0.30\ \text{m}.$$ Substituting all values,

$$B=\dfrac{9 \times 10^{-31}\; \text{kg}\; \times 6.66 \times 10^{5}\; \text{m s}^{-1}} {1.6 \times 10^{-19}\; \text{C}\; \times 0.30\ \text{m}}.$$

$$B=\dfrac{59.94 \times 10^{-26}} {0.48 \times 10^{-19}} =\dfrac{59.94}{0.48}\times 10^{-7} \approx 125 \times 10^{-7}\ \text{T}.$$

Hence, the correct answer is Option 125.