Question: Q. 8. In the study of a photoelectric effect the graph between the stopping potential V and frequency 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.

U] [Delhi II 2017]

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Solution:

Ans. (i) Q has higher threshold frequency

(ii) Work function,

ϕ0=hv0 hv0=(6.6×1034) ×6×10141.6×1019eV1/2 =2.5eV Kmax=h(vv0) =6.6×1034×2×10141.6×1019eV Kmax=0.82eV

[CBSE Marking Scheme, 2017]

Detailed Answer :

(ii) Work function of Qϕ0=hv0

(iii)=6.6×1034×6×10141.6×1019eV =2.47eV hv0=6.6×1034×8×10141.6×1019eV =3.3eV KEmax=hvhv0

(1)=(3.32.47)eV =0.83eV



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