In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 $$\mathring{A}$$. If the speed of electron is $$2.2 \times 10^6$$ m s$$^{-1}$$. Then the current associated with the electron will be _________ $$\times 10^{-2}$$ mA. [Take $$\pi$$ as $$\frac{22}{7}$$]

For a single revolving charge, current is defined as

$$I = \dfrac{\text{charge}}{\text{time period}}$$

The electron carries charge $$e = 1.6 \times 10^{-19}\,\text{C}$$.
If the electron moves in a circular orbit of radius $$r$$ with linear speed $$v$$, its time period is

$$T = \dfrac{\text{circumference}}{\text{speed}} = \dfrac{2\pi r}{v}$$

Substituting this in the current formula:

$$I = \dfrac{e}{T} = \dfrac{e v}{2\pi r} \quad\; -(1)$$

Given data:
$$r = 0.5\,\text{\AA} = 0.5 \times 10^{-10}\,\text{m}$$
$$v = 2.2 \times 10^{6}\,\text{m s}^{-1}$$
$$e = 1.6 \times 10^{-19}\,\text{C}$$
$$\pi = \dfrac{22}{7}$$

Insert the numerical values into equation $$(1)$$:

$$I = \dfrac{(1.6 \times 10^{-19})(2.2 \times 10^{6})}{2\left(\dfrac{22}{7}\right)(0.5 \times 10^{-10})}$$

Simplify the denominator first:
$$2 \left(\dfrac{22}{7}\right)(0.5 \times 10^{-10}) = \dfrac{22}{7} \times 10^{-10} = 3.142857 \times 10^{-10}$$

Simplify the numerator:
$$(1.6 \times 2.2) \times 10^{-19+6} = 3.52 \times 10^{-13}$$

Divide numerator by denominator:

$$I = \dfrac{3.52 \times 10^{-13}}{3.142857 \times 10^{-10}} = 1.12 \times 10^{-3}\,\text{A}$$

Convert amperes to milliamperes (1 mA = 10-3 A):

$$I = 1.12\,\text{mA}$$

Express $$1.12\,\text{mA}$$ in the form $$N \times 10^{-2}\,\text{mA}$$:

$$1.12\,\text{mA} = 112 \times 10^{-2}\,\text{mA}$$

Therefore, the required number is 112.