Q.26 Energy of an electron in second Bohr orbit is –3.4 eV. Calculate its kinetic energy and potential energy in third Bohr orbit.
Best answer
Answer: In the third orbit, K.E. \(\approx +1.51\) eV and P.E. \(\approx -3.02\) eV.
Energy in the \(n\)th orbit \(E_n \propto \dfrac{1}{n^2}\), so
\[E_3 = E_2 \times \frac{2^2}{3^2} = -3.4 \times \frac{4}{9} \approx -1.51\ \text{eV}\]In a Bohr orbit, K.E. \(= -E_n\) and P.E. \(= 2E_n\) (since \(E = \text{K.E.} + \text{P.E.}\), K.E. \(= \dfrac{e^2}{8\pi\varepsilon_0 r}\), P.E. \(= -\dfrac{e^2}{4\pi\varepsilon_0 r}\)).
K.E. in the third orbit \(= -E_3 = +1.51\) eV.
P.E. in the third orbit \(= 2E_3 = 2(-1.51) \approx -3.02\) eV.
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