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CBSE Class 12 Physics 2017 Delhi Set 2 Paper

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Question : 4 of 8
Marks: +1, -0
SECTION -C
In the study of a photoelectric effect the graph between the stopping potential VV and frequency vv of the incident radiation on two different metals PP and QQ 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 8×8 \times 1014 Hz10^{14} \text{ Hz} for this metal.
Solution:  
Identification of metal which has higher threshold frequency
Determination of the work function of the metal which has greater value
Calculation of maximum kinetic energy (Kmax⁡)(K_{\max}) of electron emitted by light of frequency 8×8 \times 1014 Hz10^{14} \text{ Hz}
(i) QQ has higher threshold frequency
(ii) Work function ϕ0=hv0\phi_0 = h v_0
hv0  =(6.6×10−34)×5×10141.6×10−19 eV  h v_0\;= (6.6 \times 10^{-34}) \times \frac{5 \times 10^{14}}{1.6 \times 10^{-19}} \text{ eV}\;
  =2 eV  \;=2 \text{ eV}\;
Kmax⁡  =h(v−v0)K_{\max }\;=h (v-v_0)
  =  6.6×10−34×3×10141.6×10−19 eV  \;=\;\frac{6.6 \times 10^{-34} \times 3 \times 10^{14}}{1.6 \times 10^{-19}} \text{ eV}\;
Kmax⁡  =0.83 eVK_{\max }\;=0.83 \text{ eV}
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