Light of 4.5 ev is incident on cesium surface and stopping potential is 0.25 volts so what will be maximum kinetic energy
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Explanation:
Given Light of 4.5 ev is incident on cesium surface and stopping potential is 0.25 volts so what will be maximum kinetic energy
- So we need to find the maximum kinetic energy
- V = 4.5 ev = 4.5 x 10^-19 V
- Q = 0.25 volts
- Now we have E = VQ
- So V = E/Q
- Therefore we have
- Kinetic Energy = Potential difference x Charge
- E = V x Q
- = 4.5 x 1.6 x 10^-19 x 0.25
- = 7.2 x 0.25 x 10^-19
- = 1.8 x 10^-19 Joule
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https://brainly.com/question/12544524
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Answer:
0.25ev
Explanation:
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