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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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Answer» tion: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 REFERENCE link will behttps://brainly.com/question/12544524 |
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