Question

Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.125 nm , and their final separation is 0.103 nm .

Part A

How much electric potential energy was lost in this reaction (in units of eV)?

Answer 1

here

the charge of electron = -1.6022 * 10^-19 C

the charge of proton = 1.6022 * 10^-19 C

Ui = k * q^2 / ri

Ui = 9 * 10^9 * (1.6022 * 10^-19)^2 / 0.125 * 10^-9

Ui = 1.8482 * 10^-18 J

then

Uf = k * q^2 / rf

Uf = 9 * 10^9 * (1.6022 * 10^-19 )^2 / 0.103 * 10^-9

Uf = 2.2430 * 10^-18 J

therefore

Uf - Ui = 2.2430 * 10^-18 - 1.8482 * 10^-18

Uf - Ui = 3.948 * 10^-19 J

then convert it into ev

1 J = 6.241509 * 10^18 eV

3.948 * 10^-19 * 6.241509 * 10^18 = 2.46 eV

so the electric potential energy was lost in this reaction is 2.46 eV