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 rn 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 nth  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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