Question

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A Particle carrying charge of 2e falls through a Potential difference of 3.ov
Calculate the energy acquired 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

Explanation:

Answer 2

The energy acquired by the particle is 6 eV.

Explanation:

The relation between potential difference and energy is given by

$E=q\Delta V$

Here, q is the charge and $\Delta V$ is the potential difference.

Given $q = 2 e =2\times1.6\times10^{-19} C$ and $\Delta V=3V$.

Substitute the given values, we get

$E=2\times1.6\times10^{-19} C\times3V$

$E = 6\times1.6\times10^{-19} J$

E = 6 eV.

Thus, the energy acquired by the particle is 6 eV.

#Learn More:

Topic: Potential difference.

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