Class 12th Physics 2017 Set2 Delhi Board Paper Solution
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.
(i) Einstein’s photoelectric equation
hν = φ+eV
V = hν/e + φ/e …(i)
Equation (i) represents a straight line given by line P andQ. φ/e represents negative intercept on the Y-axis. Since,Q has greater negative intercept, it will have greater φ (work function) and hence higher threshold frequency.
(ii) Also from
W = hv0 work function of Q is 6 x 1014 x 6.63 x 10-34
W = 3.9 x 10-19 unit WQ > WP
(iii) Emax = hv - φ0 = hv - hv0 = h(v-v0)
= 6.63 (8x1014 - 6x1014) J x 10-34J
= 1.33 x 10-19 J
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