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 × 10^{14} Hz for this metal.

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#### Solution

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

(ii)

Work function, w_{0}=hν_{0}

where, v_{0} is threshold frequency.

(ν_{0})Q>(ν_{0})_{P}

⇒W_{Q}>W_{P}

(iii) Maximum kinetic energy is given by

K=E−hν_{0}

W_{Q}>W_{P}

Hence the kinetic energy for metal Q is

E=hν−hν_{0}

= h(v−v0)

=6.63×10^{−34}×(8×10^{14}−6×10^{14})

=1.33×10^{−19} J

Concept: Electromagnetic Waves

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