# Radius of Hydrogen Orbit and Velocity of Electron

**Radius of Hydrogen Orbit:**

For a hydrogen atom Bohr assume that the electrostatic force of attraction between positively charged nucleus (+ze) and negatively charged electron (-e) provides the necessary amount of centripetal force for revolution.

i.e., Electrostatic force between electron and nucleus = Centripetal force

$$or, \frac{1}{4 \pi \varepsilon_\circ}\frac{ze^2}{r^2} = \frac{mv^2}{r}$$

$$or, r = \frac{1}{4 \pi \varepsilon_\circ}\frac{ze^2}{mv^2}……(i)$$

From Bohr’s Postulates,

$$mvr = \frac{nh}{2\pi}$$

$$or, r = \frac{nh}{2\pi mv}……(ii)$$

From equation (i) and (ii),

$$\frac{1}{4 \pi \varepsilon_\circ}\frac{ze^2}{mv^2} =\frac{nh}{2\pi mv}$$

$$or, v = \frac{ze^2}{2\varepsilon_\circ nh}……(iii)$$

For hydrogen atom, z =1,

$$v = \frac{e^2}{2\varepsilon_\circ nh}$$

Putting the value of V in equation (ii),

$$r = \frac{nh}{2 \pi me^2} \times 2 \varepsilon_\circ nh$$

$$\therefore r = \frac{\varepsilon_\circ n^2 h^2}{\pi me^2}$$

For nth orbit of hydrogen atom,

$$\ r _n = \frac{\varepsilon_\circ n^2 h^2}{\pi me^2}…….(iv)$$

So, Bohr’s atomic model gives the radius of hydrogen atom.

**Velocity of the electron:**

From Bohr’s postulates, we know:

$$mvr = \frac{nh}{2\pi}$$

$$or, r = \frac{nh}{2\pi mv}$$

$$or, v = \frac{nh}{2\pi rm}…….(v)$$

Now, substituting the value of r_{n} from equation (iv), we get:

$$v_n = \frac{nh}{2 \pi m}\frac{\varepsilon_\circ n^2 h^2}{\pi me^2}$$

$$\therefore v_n = \frac{e^2}{2\pi \varepsilon_\circ nh}…….(vi)$$

Equation (vi) is the velocity of the electron in the n^{th } orbit. Also from the above equation, it is found that the electrons closer to the nucleus move with higher velocity than those lying farther.

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