## Energy Level Formula :

Energy Level Formula is radius of energy level r(n)_{(m)} in meter is equal to the square of n^{2} and multiply the product of radius r(1)_{(m)} in meter. Hence the Energy level formula can be written a,

**E(n)= – 1/n² * 13.6 eV**

#### Other forms of energy level formula,

If n1 is the lower energy level and n2 is the higher energy level, the equation becomes

**ΔE = hμ = -(1/n2² – 1/n1²) 13.6 eV**

Taking negative signs inside,

**ΔE = hμ = (1/n1² – 1/n2²) 13.6 eV**

This form of energy level formula is used in case of transitions between two energy levels to find the energy of the emitted photon.

## Sample Problems :

### Example 1:

calculate the energy? In terms of electrons absorbing and emitting photons to shift energy levels, the hydrogen spectrum has the longest wavelength (n_{high}) of 4 and the shortest wavelength (n_{low}) of 2. Using the Bohr energy level formula,

### Solution :

*Given,*

n_{high} = 4

n_{low} = 2

By using the Bohr formula energy levels we get,

ΔE = (1/n^{2}_{low} − 1/n^{2}_{high})13.6eV

ΔE = (1/2^{2} − 1/4^{2})13.6eV

ΔE = (0.25 – 0.0625) *13.6eV

ΔE = 0.1875 * 13.6eV

ΔE = 2.55eV

### Example 2:

Calculate the amount of energy that has been released? Electrons in one mole of hydrogen atoms fall from the n=3 energy level to the n=2 energy level.

### Solution :

Using the following equation for the energy of an electron in Joules:

ΔE = −2.18 *10^{−18}(1/n^{2}_{f} − 1/n^{2}_{i})

And

1mol = 6.02 * 10^{23} molecules

Combining equations and plugging in values:

ΔE_{1mole} = 6.02 * 10^{23} * −2.18 * 10^{−18}(1/2^{2}−1/3^{2})

ΔE_{1mole} = −182kJ

182kJ would be released