Question

(b) A ray of ultraviolet light of wavelength
a = 3000 Å falls on a surface of a material
whose work function is 2.28 eV and ejects
an electron. What is the velocity of the
ejected photoelectron ?
​

Answer 1

Answer:

Refer to the attachment for your answer.

Answer 2

Answer:

The velocity of the photoelectron is 25.55×$10^{3}$m/s

Explanation:

a = 3000 angstrom

hγ₀ = work function = 2.28eV

• Energy  is given by hc/λ = 6.62× $10^{-34}$×3 ×$10^{8}$/3000×$10^{-10}$

                                                = 6.62×$10^{-19}$Joules

• hγ = hγ₀ + 1/2mv² (for calculating v we will substitute the values obtained)
• hc/λ = hc/λ₀ +1/2mv²

• (6.62-3.648)×$10^{-19}$ = 2.97*$10^{-19}$
• to find velocity we have to divide above answer obtained by 9.1×$10^{-31}$, we get v as follows

                       v = √657×$10^{9}$

                       v = 25.55 ×$10^{3}$m/s