Question

Order of number of revolutions per second $ {\gamma _1},\;{\gamma _2},\;{\gamma _3} $ and $ {\gamma _4} $ for I, II, III, IV orbits is:
A. $ {\gamma _1} > {\gamma _2} > {\gamma _3} > {\gamma _4} $
B. $ {\gamma _4} > {\gamma _3} > {\gamma _2} > {\gamma _1} $
C. $ {\gamma _1} > {\gamma _2} > {\gamma _4} > {\gamma _3} $
D. $ {\gamma _2} > {\gamma _3} > {\gamma _4} > {\gamma _1} $

Answer 1

Hint: According to Bohr’s theory, the orbits in which electrons are present within an atom are quantized due to which the amount of energy absorbed or released is also quantized and the change in energy of orbitals is represented $ \Delta E = hf $ , where hf is the energy of absorbed or emitted photon in form of packets known as quanta.

Complete step by step answer:
According to the Bohr concept, the radius of orbit around the nucleus and velocity of an electron present in various orbits can be calculated as follows:
According to Bohr’s first postulate:
$ \dfrac{{mv_n^2}}{{{r_n}}} = \dfrac{{KZ{e^2}}}{{r_n^2}}\;\; - (1) $
According to Bohr’s second postulate:
$ m{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}\;\;\; - (2) $
Where, $ {r_n} $ is the radius of nth orbit, $ {v_n} $ is the velocity of nth orbit, m is the mass of electron, Z is the atomic number of element, K is coulomb’s constant, h is Planck’s constant and e is the charge on the electron.
On solving the equations (1) and (2), the expressions for radius and velocity for nth orbital of an atom are as follows:
Radius r $ = 0.53\left( {\dfrac{{{n^2}}}{Z}} \right)\mathop {\text{A}}\limits^ \circ $
Velocity v $ = 2.165 \times {10^6}\left( {\dfrac{Z}{n}} \right)\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} $
We know that, time taken by an electron to complete one revolution in an orbit can be calculated as per following expressions:
$ T = \dfrac{{2\pi r}}{v} $
Substituting the values of velocity and radius calculated for nth orbit according to Bohr’s concept:
$ \Rightarrow T = 2\pi \times \left( {0.53 \times {{10}^{ - 10}} \times \dfrac{{{n^2}}}{Z}} \right) \times \dfrac{n}{{2.165 \times {{10}^6} \times Z}} $
Therefore, time of revolution of an electron varies as follows:
$ \Rightarrow T \propto \dfrac{{{n^3}}}{{{Z^2}}}\;\; - (2) $
Now, we need to find the order of number of revolutions per second i.e., order of frequency for the given orbits. We know that,
Frequency $ \nu \propto \dfrac{1}{T} $
From equation (2), variation of frequency with nth orbit and atomic number can be expressed as follows:
$ \Rightarrow \nu \propto \dfrac{{{Z^2}}}{{{n^3}}} $
According to the expression, frequency of an electron is inversely proportional to the value of orbit i.e., greater the number of orbits, lesser will be its frequency. Therefore, we can conclude that the number of revolutions per second decreases on increasing orbitals.
Hence, the correct order of number of revolutions per second is $ {\gamma _1} > {\gamma _2} > {\gamma _3} > {\gamma _4} $ . Thus, option (A) is the correct.

Note:
It is important to note that the Bohr model is only applicable to hydrogen atoms or atoms like hydrogen atoms. It was the major limitation of Bohr’s atomic model. In recent times, the quantum mechanical model is preferred over Bohr’s model because it is a better version which is based on a solution to Schrodinger’s equation.