Question: The kinetic energy of an electron in the second Bohr’s orbit of a hydrogen atom is: [${a_0}$ is the Bohr’s radius]
(A) $\dfrac{{{h^2}}}{{4{\pi ^2}m{a_0}^2}}$
(B) $\dfrac{{{h^2}}}{{16{\pi ^2}m{a_0}^2}}$
(C) $\dfrac{{{h^2}}}{{32{\pi ^2}m{a_0}^2}}$
(D) $\dfrac{{{h^2}}}{{64{\pi ^2}m{a_0}^2}}$

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Hint: Before attempting this question one should have prior knowledge of Bohr model, working on it and basic formulas given by Bohr’s model, using the given information will help you to approach towards the solution to the question.

Complete step by step solution:
Niels Bohr introduced an atomic hydrogen model, in which he describes a positively charged nucleus, inside which are protons and neutrons which are then surrounded by negatively charged electron clouds. Since, Bohr describes that the electron orbits the nucleus in an atomic shell. Hence, according to the question we are asked the kinetic energy of an electron in the second Bohr’s orbit of a hydrogen atom. So, according to Bohr, the angular momentum is given as follows.
$mvr = \dfrac{{nh}}{{2\pi }}$
From here, mv=$\dfrac{{nh}}{{2\pi r}}$ (equation 1)
Since, we have to find out the kinetic energy,
So kinetic energy =$\dfrac{1}{2}m{v^2}$ (equation 2)
By equating these two equations we get,
$(mv) = \dfrac{{nh}}{{2\pi r}}$
By squaring the above equation we get,
$(m{v^2}) = \dfrac{{{n^2}{h^2}}}{{({2^2}){\pi ^2}{r^2}}}$
Which is then equal to ${m^2}{v^2} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{r^2}}}$
Now, let us move m towards left side then, we will get the equation as:
$m{v^2} = \dfrac{1}{m} \times \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{r^2}}}$ (equation 3)
Now, let us put equation 3 in equation 1
Then, kinetic energy = $\dfrac{1}{2} \times \dfrac{1}{m} \times \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{r^2}}}$
Now, as we have given in the question that radius of the Bohr’s=${a_0}$ and the electron is in the second shell, so n=2 and ${r_1} = {a_0}$ (given). As the electron is in the second shell Thus, ${r_2} = {a_0} \times {(2)^2} = 4{a_0}$
Let us put the final values to the above equation:
$K.E = \dfrac{1}{2} \times \dfrac{{{2^2}{h^2}}}{{4{\pi ^2}{{(4{a_0})}^2}m}}$
=$\dfrac{{{h^2}}}{{32{\pi ^2}{a_0}^2m}}$

Hence, Option C is correct.

Note: As we all know, the Bohr’s model was an important step in development of atomic theory, but it has several limitations too which are as follows:
According to Bohr’s model the electrons have both a known radius and orbit, but this concept causes violation of the Heisenberg uncertainty principle which says that it is impossible to simultaneously determine the momentum and position of an electron.
The Bohr model does not predict the spectra when larger atoms are in question.
It also cannot predict the intensities of several lines.
The Bohr model does not tell us about the fact that when an electron is in acceleration, it emits electromagnetic radiation.

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