Question

When 7.9×1014 Hz light shines on a plate of unknown material, it is determined that the stopping potential is 1.1 V. Determine the work function (in eV) & the cutoff frequency. Wo = eV fo =

Answer 1

W = 2.158 eV

fo = 5.23 x 10^14 Hz

Step-by-step explanation:

f = 7.9 x 10^14 Hz, Vo = 1.1 V

Let W be the work function.

Use the Einstein equation

Energy = W + eVo

hf = W + eVo

where, h is the Plank's constant and e be the electronic charge.

W = hf - eVo

W = (6.6 x 10^-34 x 7.9 x 10^14) - (1.6 x 10^-19 x 1.1)

W = 5.214 x 10^-19 - 1.76 x 10^-19 = 3.454 x 10^-19 J

W = (3.454 x 10^-19) / (1.6 x 10^-19) = 2.158 eV

Let fo be the cut off frequency

W = h fo

fo = W / h = (3.454 x 10^-19) / (6.6 x 0^-34) = 5.23 x 10^14 Hz