Question

A particle carrying a charge of 2e falls through a potential difference of 3V. The energy required by it

Answer 1

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

Answer 2

Answer:

here is the solution in the pic

Explanation:

I think it will help you for sure