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Grade: 12
        The wave length of an incident light is 6000A° if work function of the material is 1ev calculate the velocity of the emitted electron
7 months ago

Answers : (1)

Arun
12304 Points
							
Dear Pooja
 
E = Wo + Kmax
 
E = 12375/6000 = 2.0625 ev
 
Now
Kmax = E - Wo = 2.0625 - 1 = 1.0625 ev
 
Now
1/2* m* V² = 1.0625 * 1.6 * 10^-19 joules
 
Put m = 9.1 * 10^-31 kg
 
And find V
 
 
Regards
Arun (askIITians forum expert)
6 months ago
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