Question

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.50×1015Hz1.50×1015Hz ? Express your answer in joules to three significant figures.

Answer 1

Kinetic Energy = 6.51x10^-19 J (3 s.f)

Step-by-step explanation:

frequency = 1.50×1015 s^-1

Kinetic Energy = ?

Kinetic Energy = ħf - ф

h = plank's constant = 6.626x10^-34 Js

hf = 6.626x10^-34 * 1.50×10^15 = 9.939x10^-19 J

Work function, ф = 2.1 eV = 3.3642x10^-19 J (upon conversion to joules

Kinetic energy = 9.939x10^-19 J - 3.3642x10^-19 J

Kinetic Energy = 6.511x10^-19 J

Kinetic Energy = 6.51x10^-19 J (3 s.f)