Let an imaginary atom have only one electron in an orbit at a distance of \(1\,\mathring{\mathrm{A}}\) from one proton in the nucleus. The potential energy of the system will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-23.04\times10^{-19}\,\mathrm{J}\)
The potential energy of two point charges is the work done in bringing them together from infinity. For an electron and a proton the charges are opposite, so the energy is negative.
\( \displaystyle U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r} \)
\( \displaystyle r = 1\,\mathring{\mathrm{A}} = 10^{-10}\,\mathrm{m} \)
\( \displaystyle U = \frac{(9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}) \times (1.6\times10^{-19}\,\mathrm{C}) \times (-1.6\times10^{-19}\,\mathrm{C})}{10^{-10}\,\mathrm{m}} \)
\( \displaystyle U = -23.04\times10^{-19}\,\mathrm{J} \)