Question

What is the frequency of revolution of electrons present in $ {2^{nd}} $ Bohr's orbit of H-atom?
(A) $ 8.5 \times {10^6}Hz $
(B) $ 8.5 \times {10^{16}}Hz $
(C) $ 7.82 \times {10^{15}}Hz $
(D) $ 8.13 \times {10^{16}}Hz $

Answer 1

Hint :In this problem we’ll make use of the Bohr’s Model and use the formulas for finding radius and velocity for the Hydrogen and Hydrogen like atoms. Bohr’s model is said to be semi classical as it combines both classical concepts as well as quantization concepts.

Complete Step By Step Answer:
We are given a Hydrogen atom which has only one electron. Bohr's theory is applicable to Hydrogen and Hydrogen like species (like $ H{e^ + },L{i^{ + 2}} $ ,etc) . The radius of the atom can be given by the formula:
  $ {r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}}} \times \dfrac{1}{Z} $
This can be simplified as: $ {r_n} = 0.529 \times \dfrac{{{n^2}}}{Z}{A^ \circ } $
For the given Hydrogen atom where the atomic number, $ z = 1 $ the formula becomes $ {r_n} = 0.529 \times {n^2} $
Here, $ {r_n} $ is the radius of the $ {n^{th}} $ Bohr orbit. n is the Bohr Orbit number in which the electron is present (here $ n = 2 $ ) and Z is the atomic number of the element.
The velocity of an electron in the Bohr orbit is given by a similarly simplified formula.
  $ v = 2.18 \times {10^6} \times \dfrac{z}{n} $
For hydrogen atom with atomic number one, the formula becomes $ {v_n} = 2.18 \times {10^6} \times \dfrac{1}{n} $
The time taken for the electron in a Bohr’s Orbit is given as $ T = \dfrac{{2\pi r}}{v} $
Where r is the radius of the Bohr’s orbital and v is the velocity of the electron. Substituting the values from r and v from the equation, the time T can be given as:
  $ T = 1.5211 \times {10^{ - 16}} \times \dfrac{{{n^3}}}{{{Z^2}}}\sec $
The frequency is the inverse of time taken. Hence the frequency can be given as:
  $ F = \dfrac{1}{T} = 6.55 \times {10^{15}} \times \dfrac{{{Z^2}}}{{{n^3}}} $
Substituting the values of Z and n as $ Z = 1{\text{ and }}n = 2 $ . The frequency can be given as:
 $ F = 6.55 \times {10^{15}} \times \dfrac{{{1^2}}}{{{2^3}}} = 6.55 \times {10^{15}} \times \dfrac{1}{8} $
  $ F = 8.13 \times {10^{16}} $ Hz
Hence the correct option is Option (d).

Note :
If the frequency and Radius were already given then we can directly use the formula $ f = \dfrac{v}{{2\pi r}} $ to find the frequency of the electron. Also this formula is applicable to hydrogen and hydrogen like species only.