Answer :

According to Rutherford’s model of an atom, for dynamic stability of an atom the electronic forces on the electron must account for the centripetal forces acting on it, thus,

Where, m_{e} is the mass of the electron,

v_{n} is the velocity of the electron in the n^{th} shell

r_{n} is the radius of the orbit

F_{e} is the electrostatic force between revolving electron and nucleus

F_{c} is the centripetal force required by the electron to move in circular orbits of radius r

By coulomb’s law we can write,

For a hydrogen atom Z=1,

The kinetic energy (K) is,

The Electrostatic potential energy (U) is,

(The negative sign in U signifies that the electrostatic force is in the –r direction.) Thus, the total energy E of the electron in a hydrogen atom is,

The total energy being negative signifies that the electron is bound to the nucleus.

**OR**

Suppose m be the mass of an electron and v_{n} be its speed in the n^{th} orbit of radius r_{n}. From Rutherford model, the centripetal force for revolution is produced by electrostatic attraction between electron and nucleus.

For dynamic stability of an atom the electronic forces on the electron must account for the centripetal forces acting on it, thus,

Where, m_{e} is the mass of the electron, v_{n} is the velocity of the electron in the n^{th} shell and r_{n} is the radius of the orbit

By coulomb’s law we can write,

For a hydrogen atom Z=1,

Bohr’s model of atom postulates that the values of angular momentum of the electron should be the integral multiples of h/2, so,

where n is an integer, r_{n} is the radius of nth possible orbit and v_{n} is the

speed of moving electron in the nth orbit.

Using the above relation and equation (a) we get,

no

The Expression of Bohr’s radius can be obtained by substituting n=1,

Substituting the values of all the constants,

We get the value Bohr’s radius as, **a _{0} = 5.29 × 10^{-11} m**

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Using Rutherford Physics - Board Papers