मराठी

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 - Physics

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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

  1. the energy of the incident radiation,
  2. the maximum kinetic energy of the emitted electron and
  3. the stopping potential of the surface.
संख्यात्मक
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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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