Using Rutherford model of the atom, derive ...

CBSE Class 12 Physics 2014 Outside Delhi Set 1 Paper

Question : 9 of 30

Marks: +1, -0

Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?

Using Bohr's postulates of the atomic model, derive the expression for radius of

n^{\text{th}}

electron orbit. Hence, obtain the expression for Bohr's radius.

Solution:

According to Rutherford,

\frac{m v^2}{r} = \frac{1}{4\pi \varepsilon_0} \frac{Z e^2}{r^2}

m v^2 = \frac{1}{4\pi \varepsilon_0} \frac{Z e^2}{r}

\text{Total energy} = \text{P.E.} + \text{K.E.}

= -\frac{1}{4\pi \varepsilon_0} \frac{Z e^2}{r} + \frac{1}{8\pi \varepsilon_0} \frac{Z e^2}{r}

= -\frac{1}{8\pi \varepsilon_0} \frac{Z e^2}{r}

The negative sign shows that electron-nucleus form an abound system.

According to Bohr's model, electrons revolve around the orbits such that their angular momentum is an integral multiple of

\frac{h}{2\pi}

m v r = \frac{n h}{2\pi}

\frac{m v^2}{r} = \frac{1}{4\pi \varepsilon_0} \frac{Z e^2}{r^2}

r = \frac{\varepsilon_0 n^2 h^2}{\pi Z e^2 m}

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