# How would you calculate the energy released as a photon drops from n=3 to the ground state?

May 13, 2017

$\Delta E$ =-1.936×10^-18J

#### Explanation:

$\Delta E$ = $K$ $\left(\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right)$
Where ${n}_{1}$ =$1$ and ${n}_{2}$ =3, since the photon falls from excited to ground state the energy is released.

$\Delta E$ = $K$ $\left(\frac{1}{1} ^ 2 - \frac{1}{3} ^ 2\right)$
$\Delta E$ = $K$ $\left(\frac{1}{1} - \frac{1}{9}\right)$
But $K$ = -2.178×10^-18J, a constant value for calculating energy of an electron in H atom
$\Delta E$ = -2.178×10^-18$\left(\frac{1}{1} - \frac{1}{9}\right)$
$\Delta E$ =-1.936×10^_18J