Question

A light with frequency of 3.5 x 10^15 Hz is utilized to illuminate the selenium metal (work function 5.11 eV). What will be the maximum velocity of the ejected photoelectrons in 10^6 m/sec?

Answer 1

Answer:

The maximum speed of the ejected photoelectrons is 1.815 x 10⁶ m/s.

Explanation:

Given;

frequency of the light, f = 3.5 x 10¹⁵ Hz

work function of the metal, Φ = 5.11 eV

                                              Φ = 5.11 x 1.602 x 10⁻¹⁹ J = 8.186 x 10⁻¹⁹ J

The energy of the incident light is given as sum of maximum kinetic energy and work function of the metal.

E = K.E + Φ

where;

E is the energy of the incident light, calculated as;

E = hf

E = (6.626 x 10⁻³⁴)(3.5 x 10¹⁵)

E = 2.319 x 10⁻¹⁸ J

The maximum kinetic energy of the photoelectrons is calculated as;

K.E = E - Φ

K.E = 2.319 x 10⁻¹⁸ J  - 8.186 x 10⁻¹⁹ J

K.E =  2.319 x 10⁻¹⁸ J - 0.8186 x 10⁻¹⁸ J

K.E = 1.5004 x 10⁻¹⁸J

The maximum speed of the ejected photoelectrons in 10⁶ m/s  is given as;

K.E = ¹/₂mv²

$v_{max}^2 = \frac{2K.E}{m} \\\\v_{max}= \sqrt{\frac{2K.E}{m}} \\\\v_{max} = \sqrt{\frac{2(1.5 \ \times \ 10^{-18})}{(9.11 \ \times \ 10^{-31})}}\\\\v_{max} =1.815 \ \times \ 10^{6} \ m/s$

Therefore, the maximum speed of the ejected photon-electrons is 1.815 x 10⁶ m/s.