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Lemur [1.5K]
3 years ago
9

What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?

Physics
1 answer:
Vlada [557]3 years ago
5 0

Answer:

Ep=-2.3*10^{-19}J

Explanation:

The change in potential energy can be expressed as:

Ep=K.\frac{q1.q2}{r}

where K is a constant with a value of 9*10^{9}\frac{N.m^{2}}{C^{2}}, q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is +1.6*10^{-19}C and the charge for the electron is -1.6*10^{-19}C.

Converting r=1.0nm to m:

1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m

Replacing values:

Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}

Ep=-2.3*10^{-19}J

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