Question

A photoelectric-effect experiment finds a stopping potentialof 1.93V when light of 200nm is used to illuminate thecathode.

a) From what metal is the cathode made from?
b) What is the stopping potential if the intensity of thelight is doubled?

Answer 1

a) Tantalum

b) 1.93 V

Step-by-step explanation:

The energy of the incident photon= hc/λ

h= Plank's constant=6.63×10^-34 Is

c= speed of light = 3×10^8 ms-1

λ= wavelength of incident photon

E= 6.63×10^-34 × 3×10^8/ 200×10^-9

E= 0.099×10^-17

E= 9.9×10^-19 J

The kinetic energy of the electron = eV

Where;

e= electronic charge = 1.6×10^-19 C

V= 1.93 V

KE= 1.6×10^-19 C × 1.93 V

KE= 3.1 ×10^-19 J

From Einstein's photoelectric equation;

KE= E -Wo

Wo= E -KE

Wo=9.9×10^-19 J - 3.1 ×10^-19 J

Wo= 6.8×10^-19 J

Wo= 6.8×10^-19 J/1.6×10^-19

Wo= 4.25 ev

The metal is Tantalum

b) the stopping potential remains 1.93 V because intensity of incident photon has no effect on the stopping potential.