Question

In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 Å. If the speed of the electron is $2.2 \times 10^6$ m/s, then the current associated with the electron will be _________ $\times 10^{-2}$ mA. [Take $\pi$ as 22/7]

Hint:

The concept of an orbiting charge creating a current is fundamental. The equivalent current is always $I=q/T = qf$, where q is the charge, T is the period, and f is the frequency of revolution. This is also related to the magnetic dipole moment of the orbit.

Answer 1

Correct Answer: 112

An electron revolving in an orbit constitutes a current. The magnitude of the current I is the charge e divided by the time period T of one revolution.
$I = \frac{e}{T}$
The time period T is the circumference of the orbit divided by the speed v of the electron.
$T = \frac{2\pi r}{v}$
Substituting this into the current equation:
$I = \frac{e}{\left(\frac{2\pi r}{v}\right)} = \frac{ev}{2\pi r}$
We are given the following values in SI units:
Charge of electron, $e = 1.6 \times 10^{-19}$ C.
Speed of electron, $v = 2.2 \times 10^6$ m/s.
Radius of orbit, $r = 0.5$ Å = $0.5 \times 10^{-10}$ m.
$\pi = 22/7$.
$I = \frac{(1.6 \times 10^{-19}) \times (2.2 \times 10^6)}{2 \times \frac{22}{7} \times (0.5 \times 10^{-10})}$
$I = \frac{1.6 \times 2.2 \times 10^{-13}}{1 \times \frac{22}{7} \times 10^{-10}} = \frac{3.52 \times 10^{-13}}{\frac{22}{7} \times 10^{-10}}$
$I = \frac{3.52 \times 7}{22} \times 10^{-3} = \frac{24.64}{22} \times 10^{-3} = 1.12 \times 10^{-3}$ A.
The current is $1.12 \times 10^{-3}$ A, which is equal to 1.12 mA.
The question asks for the answer in the form of $x \times 10^{-2}$ mA.
$1.12 \text{ mA} = x \times 10^{-2} \text{ mA}$
$x = \frac{1.12}{10^{-2}} = 1.12 \times 100 = 112$.