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calculate kinetic energy of ejected electron when uv-radiations of frequency 1.6*10^15 strikes the surface of potassium metal .threshold frequency is 5*10^14
calculate kinetic energy of ejected electron when uv-radiations of frequency 1.6*10^15 strikes the surface of potassium metal .threshold frequency is 5*10^14

```
2 years ago

Arun
25768 Points
```							Frequency of uv light, fy = 1.6×10^15 HzThreshold frequency of Potassium, f0 = 5×10^14 HzSince the frequency (and hence energy) of the incident light is more than the threshold frequency, electron will be emitted from its surface. The amount of energy which will be used to remove the electron is called work function, given by (h×f0)The rest amount of energy of the incident light will be used to provide kinetic energy to the emitted electron.So KE of electron = energy of incident light - work function of potassiumOr KE = h(fy) - h(f0)Or KE = h×( fy- f0)Or KE = 6.63×10^(-34) × (1.6×10^15 - 5×10^14) Or KE = 6.63×10^(-20) × 11Or KE = 6.793 × 10^(-19) JKE of emitted electron is 6.793 × 10^(-19) J
```
2 years ago
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