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Question: Answered & Verified by Expert
According to the Bohr's theory of hydrogen atom, the speed of the electron, energy and the radius of its orbit vary with the principal quantum number $\mathrm{n}$, respectively, as
PhysicsCurrent ElectricityVITEEEVITEEE 2010
Options:
  • A $\frac{1}{\mathrm{n}}, \frac{1}{\mathrm{n}^{2}}, \mathrm{n}^{2}$
  • B $\frac{1}{\mathrm{n}}, \mathrm{n}^{2}, \frac{1}{\mathrm{n}^{2}}$
  • C $\mathrm{n}^{2}, \frac{1}{\mathrm{n}^{2}}, \mathrm{n}^{2}$
  • D $\mathrm{n}, \frac{1}{\mathrm{n}^{2}}, \frac{1}{\mathrm{n}^{2}}$
Solution:
1879 Upvotes Verified Answer
The correct answer is: $\frac{1}{\mathrm{n}}, \frac{1}{\mathrm{n}^{2}}, \mathrm{n}^{2}$
According to Bohr's theory of hydrogen atom,
(i) The speed of the electron in the $n$th orbit
$$
V_{n}=\frac{1}{n} \frac{e^{2}}{4 \pi \varepsilon_{0}(h / 2 \pi)} \text { or } v_{n} \propto \frac{1}{n}
$$
(ii) The energy of the electron in the $n$th orbit
$$
E_{n}=\frac{m e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}=\frac{-13.6}{n^{2}} e V \text { or } E_{n} \propto \frac{1}{n^{2}}
$$
(iii) The radius of the electron in the $n$th orbit
$$
r_{n}=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m e^{2}}=n^{2} a_{0} \text { or } r_{n} \propto n^{2}
$$
where $a_{0}=\frac{h^{2} \varepsilon_{0}}{\pi m e}=5.29 \times 10^{-11} \mathrm{~m}$

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