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Grade: 11

                        

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

Answers : (1)

Arun
24742 Points
							
Frequency of uv light, fy = 1.6×10^15 Hz
Threshold frequency of Potassium, f0 = 5×10^14 Hz
Since 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 potassium
Or 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) × 11
Or KE = 6.793 × 10^(-19) J

KE of emitted electron is 6.793 × 10^(-19) J
2 years ago
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