CBSE Class 12 · 2017 · Delhi · Modern PhysicsShort Answer

In the study of the photoelectric effect, the graph between the stopping potential

V

and frequency

ν

of the incident radiation on two different metals P and Q is shown.
(i) Which one of the two metals has the higher threshold frequency?
(ii) Determine the work function of the metal which has the greater value.
(iii) Find the maximum kinetic energy of the electron emitted by light of frequency

8 \times 1 0^{14} Hz

for this metal.

Answer: Q higher threshold; $ϕ_{0} \approx 2.5$ eV; $K_{ma x} \approx 0.83$ eV.

Solution

From the graph, metal P has threshold frequency

\approx 3 \times 1 0^{14} Hz

and metal Q has threshold frequency

\approx 6 \times 1 0^{14} Hz

.
(i) Metal Q has the higher threshold frequency.
(ii) Its work function:

ϕ_{0} = h ν_{0} = (6.6 \times 1 0^{- 34}) (6 \times 1 0^{14}) = 3.96 \times 1 0^{- 19} J \approx 2.5 eV .

(iii) For

ν = 8 \times 1 0^{14} Hz

K_{ma x} = h (ν - ν_{0}) = 6.6 \times 1 0^{- 34} (8 - 6) \times 1 0^{14} = 1.32 \times 1 0^{- 19} J \approx 0.83 eV .

Q; ϕ_{0} \approx 2.5 eV; K_{ma x} \approx 0.83 eV

CBSE Class 12 Board · 2017 · Delhi · Subjective

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CBSE Class 12 · 2017 · Delhi · Modern PhysicsShort Answer

Solution

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