# Energy of the Electron in nth Orbit

When electron is revolving around the electron, it has kinetic energy. Since, it is attracted towards the nucleus by electrostatic force of attraction, it has potential energy. Therefore, the sum of kinetic energy and potential energy gives the total energy.

We know, Kinetic energy of electron in nth orbit is given by:
$$KE = \frac{1}{2}m{v_n}^2……..(i)$$

We know from previous topic, $$v_n = \frac{e^2}{2\varepsilon_\circ nh}$$
So, putting the value of Vn in equation (i), we get:
$$KE = \frac{1}{2}m\left ( \frac{e^2}{2\varepsilon_\circ nh}\right )^2$$
$$\therefore KE = \frac{me^4}{8{\varepsilon_\circ}^2n^2h^2}$$

Similarly, Potential energy (P.E.) = Potential due to nucleus at distance r × charge of electron i.e.,

$$PE=\frac{e}{4\pi\varepsilon_\circ r_n}\times (-e)$$
$$or, PE=-\frac{1}{4\pi\varepsilon_\circ}\frac{e^2}{r_n}$$

We know from the previous topic, $$\ r _n = \frac{\varepsilon_\circ n^2 h^2}{\pi me^2}$$

Then Potential energy becomes,

$$PE=-\frac{e^2}{4\varepsilon_\circ}\frac{\pi me^2}{\varepsilon_\circ n^2 h^2}$$

$$\therefore PE=-\frac{me^4}{4{\varepsilon_\circ}^2n^2h^2}$$

$$\text{Now, Total Energy}(E_n) = \text{Kinetic Energy + Potential Energy}$$
$$or, E_n = \frac{me^4}{8{\varepsilon_\circ}^2n^2h^2} – \frac{me^4}{4{\varepsilon_\circ}^2n^2h^2}$$
$$\therefore E_n = \frac{me^4}{8{\varepsilon_\circ}^2n^2h^2}$$

Putting the values of m, e, and h in above equation as;

m = 9.1×10-31 ­­kg

e = 1.6×10-19 C

εo = 8.85×10-12 F/m

h = 6.62×10-34Js we get,

∴ Total Energy of nth orbit = – 13.6/n2 eV…….(iii)

The equation (ii) is a expression for the energy of the electron in nth orbit. The negative sign shows that the electron is bound to the nucleus. This also suggests that the energy of an electron is always negative.

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#### 4 Responses to “Energy of the Electron in nth Orbit”

1. Atul singh oberoi

Why energy of an electron in nth orbit is negative?

• Nivitha

negative sign indicated that the electron is under attraction of nucleus of is not free.

2. S.AYESHA

good

3. S.AYESHA

nice solution given to me satisfactory