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CaA Particle carrying charge of 2e falls through a Potential difference of 3.ovCalculate the energy acquired by it?​

Answer»

ANSWER: Charge on particle = 2e

Potential difference = 3.0 V

Energy?

By applying FORMULA

1/2 mv²= QV

K. E = 2e×3

= 2×1.6×10^-19×3

Energy = 9.6×10^-19J

Hope you get it CLEARLY, Thanks

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