. The threshold frequency v0 for a metal


is 7.0 × 1014  s-1. Calculate the kinetic energy of an electron emitted when radiation of frequency v = 1.0 × 1015 s-1hits the metal.

Best Answer

Explanation:

According to Einstein's equation,

Kinetic energy = 1/2mev2= h( v-v0)

h = Planck’s constant = 6.626×10-34J s

v = Frequency 

v0 =  Threshold frequency

= (6.626×10-34J s)(1.0×1015 s-1 − 7.0×1014  s-1)

= (6.626×10-34 Js)(10.0×1014 − 7.0×1014)

= (6.626×10-34 Js)(3.0×1014s-1)

= 1.988×10-19J

Final Answer:

Hence the  kinetic energy of an electron emitted is 1.988×10−19 J.

 

 

 

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