An electron moves at right angle to a magnetic field of 1.5×10^-2 T with a speed of 6 x 10⁷ m / s. If the specific charge of the electron is 1.7x 10¹¹ C/ kg. The radius of the circular path will be

2.9 cm

3.9 cm

2.35 cm

2 cm

Solution

2.35 cm

The formula for radius of circular path is

$r = \frac{m ν}{e B} = \frac{ν}{(\frac{e}{m}) B} Suppose the cut off wavelength is represented by λ_{o} so, \frac{h c}{λ_{o}} = work function = W_{o} \frac{h c}{λ_{o}} = 4.2 \times 1.6 \times 10^{- 19} λ_{o} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{4.2 \times 1.6 \times 10^{- 19}} = 2946 \times 10^{- 10} \approx 2950 A^{o}$