Q: The Photoelectric threshold wavelength of silver is 3250 × 10^{-10} m . The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10^{-10} m is :

(a) 0.6 × 10^{6} m/s

(b) 61 × 10^{3} m/s

(c) 0.3 × 10^{6} m/s

(a) 6 × 10^{4} m/s

Ans: (a)

$\large \phi = \frac{1242}{325} = 3.82eV$

$\large E = \frac{1242}{253.6} = 4.89 eV$

E = φ + K_{max}

K_{max} = E – φ = 1.077 eV

$\large \frac{1}{2}m v_{max}^2 = 1.077 \times 1.6 \times 10^{-19} $

$\large v_{max} = (\frac{2 \times 1.077 \times 1.6 \times 10^{-19} }{9.1 \times 10^{-31}})^{1/2}$

v_{max} = 0.6 × 10^{6} m/s