Q.22 An electron in hydrogen atom stays in its second orbit for 10\(^{-8}\) s. How many revolutions will it make around the nucleus in that time? [Given : e = 1.6 \(\times\) 10\(^{-19}\) C, m = 9.1 \(\times\) 10\(^{-31}\) kg]
Best answer
Answer: about \(8.2 \times 10^6\) revolutions
In Bohr’s model the period of revolution of the electron in the \(n\)th orbit is
\[T_n = \frac{4\varepsilon_0^2 h^3 n^3}{m e^4}\]so the number of revolutions per second is \(f_n = \dfrac{1}{T_n} = \dfrac{m e^4}{4\varepsilon_0^2 h^3 n^3}\). For \(n = 2\):
\[f_2 = \frac{9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^4}{4 \times (8.85 \times 10^{-12})^2 \times (6.63 \times 10^{-34})^3 \times 8}\]Numerator \(= 9.1 \times 10^{-31} \times 6.554 \times 10^{-76} = 5.964 \times 10^{-106}\).
Denominator \(= 4 \times 7.832 \times 10^{-23} \times 2.914 \times 10^{-100} \times 8 = 7.304 \times 10^{-121}\).
\[f_2 = \frac{5.964 \times 10^{-106}}{7.304 \times 10^{-121}} \approx 8.165 \times 10^{14} \text{ rev/s}\]Number of revolutions in \(t = 10^{-8}\) s:
\[N = f_2\,t = 8.165 \times 10^{14} \times 10^{-8} \approx 8.2 \times 10^6\]Log in or create a free account to answer this question.
