Caesium has a work function of 3 x 10\(^{-19}\)J. The maximum energy of liberated electrons when it is illuminated by light of frequency 6.7 x 10\(^{14}\)
HZ is? [h = 6.6 x 10\(^{-34}\)Js]
A
1.42 x 10-19J
B
3.00 x 10-19J
C
4.42 x 1010-19J
D
7.42 x 10-19J
Verified Correct Answer: A.
Physical Principles & Numerical Solution
Max kinetic energy = Energy of incident electron - work function. EK = hf - w
= 6.6 x 10-34 x 6.7 x 1014 - 3 x 10-19
= 4.422 x 10-19 - 3 x 10-19
= 1.422 x 10-19J
Source: WAEC / JAMB / NECO Classroom Archive (Record #14081)