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प्रश्न
The work function of a metal is 2.31 eV. Photoelectric emission occurs when the light of frequency 6.4 × 1014 Hz is incident on the metal surface. Calculate
- the energy of the incident radiation,
- the maximum kinetic energy of the emitted electron and
- the stopping potential of the surface.
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उत्तर
(i) Frequency of incident radiation = ν = 6.4 × 1014 Hz
Energy of incident radiation = E = hν = 6.6 × 10−34 × 6.4 × 1014 = 42.24 × 10−20 J
(ii) KEmax = hν − Φ0
∴ KEmax = 42.24 × 10−20 − 2.31 × 1.6 × 10−19 = 5.28 × 10−20 J
(iii) If stopping potential = VS, then
eVS = KEmax
∴ VS = `"KE"_"max"/"e"`
= `(5.28 xx 10^-20)/(1.6 xx 10^-19)`
= 3.3 × 10−1
= 0.33 V
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