Question

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is: ($a \circ$ is bohr radius)
A.$\dfrac{{\mathop h\nolimits^2 }}{{\mathop {4\pi }\nolimits^2 \mathop {ma \circ }\nolimits^2 }}$
B.$\dfrac{{\mathop h\nolimits^2 }}{{\mathop {16\pi }\nolimits^2 \mathop {ma \circ }\nolimits^2 }}$
C.$\dfrac{{\mathop h\nolimits^2 }}{{\mathop {32\pi }\nolimits^2 \mathop {ma \circ }\nolimits^2 }}$
D.$\dfrac{{\mathop h\nolimits^2 }}{{\mathop {64\pi }\nolimits^2 \mathop {ma \circ }\nolimits^2 }}$

Answer 1

In the Bohr model of an atom, the electrons travel in the circular orbits around the nucleus that are defined. These orbits are labelled as the quantum number n. The electrons jump from one orbit to another by emitting energy.
The formula for calculating the energy of electron is –
$\mathop E\nolimits_n = - \dfrac{{13.6}}{{\mathop n\nolimits^2 }}eV$
where,
$\mathop E\nolimits_n$= energy of the electron
$\mathop n\nolimits^2$= orbits in which the electrons travel

Complete step by step answer:
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 an follows:
$mvr = \dfrac{{nh}}{{2\pi }}$
From here, $mv = \dfrac{{nh}}{{2\pi r}}.....eqn1$
Since, we have to find out the kinetic energy,
So, kinetic energy$= \dfrac{1}{2}\mathop {mv}\nolimits^2 ....eqn2$
By equating these two equations we get,
$(mv) = \dfrac{{nh}}{{2\pi r}}$
By squaring the above equation, we get,
$\mathop {mv}\nolimits^2 = \dfrac{{\mathop n\nolimits^2 \mathop h\nolimits^2 }}{{\mathop 2\nolimits^2 \mathop \pi \nolimits^2 \mathop r\nolimits^2 }}$
Which is then equal to $\mathop {\mathop m\nolimits^2 v}\nolimits^2 = \dfrac{{\mathop n\nolimits^2 \mathop h\nolimits^2 }}{{\mathop {4\pi }\nolimits^2 \mathop r\nolimits^2 }}$
now, let us move m towards left side then, we will get the equation as:
$\mathop {mv}\nolimits^2 = \dfrac{1}{m} \times \dfrac{{\mathop n\nolimits^2 \mathop h\nolimits^2 }}{{\mathop {4\pi }\nolimits^2 \mathop r\nolimits^2 }}........eqn3$
Now, let us put equation 3 in equation 1
Then, kinetic energy = $\dfrac{1}{2} = \dfrac{1}{m} \times \dfrac{{\mathop n\nolimits^2 \mathop h\nolimits^2 }}{{\mathop {4\pi }\nolimits^2 \mathop r\nolimits^2 }}$
Now, as we have given in the question that radius of the Bohr’s$= a \circ$ and the electron is in the second shell,
So, $n = 2and\mathop r\nolimits_1 = a \circ$(given)
As the electron is in the second shell thus,
$\mathop r\nolimits_2 = a \circ \times \mathop 2\nolimits^2 = 4a \circ$
Let us put the final values to the above equation:
$K.E = \dfrac{1}{2} \times \dfrac{{\mathop 2\nolimits^2 \mathop h\nolimits^2 }}{{\mathop {4\pi }\nolimits^2 \mathop {(4a \circ )}\nolimits^2 m}} = \dfrac{{\mathop h\nolimits^2 }}{{\mathop {32\pi }\nolimits^2 \mathop {a \circ }\nolimits^2 m}}$
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 electron 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.