Question

In the study of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown below:

(i) Which one of the two metals has higher threshold frequency?

(ii) Determine the work function of the metal which has greater value.

(iii) Find the maximum kinetic energy of electron emitted by light of frequency 8 × 10¹⁴ Hz for this metal.

Answer 1

So the threshold frequency of metal Q is greater than metal P.

(ii)
Work function, w₀=hν₀

where, v₀ is threshold frequency.

(ν₀)Q>(ν₀)_P

⇒W_Q>W_P

(iii) Maximum kinetic energy is given by

K=E−hν₀

W_Q>W_P

Hence the kinetic energy for metal Q is

E=hν−hν₀

= h(v−v0)

=6.63×10⁻³⁴×(8×10¹⁴−6×10¹⁴)

=1.33×10⁻¹⁹ J