Question

Question 5 of 9 Submit What is the energy of light that must be absorbed by a hydrogen atom to transition an electron from n = 3 to n = 6? (1 2 3 +/- 0 x 100 Tap here or pull up for additional resources

Answer 1

The energy of the electron is given by:

$E_{n} = -\frac{2.18 * 10^{-18}}{n^{2}}$

Where n is the principle quantum number.

We need to find the difference between the energy of the electron in the n=3 and n=6 levels

The difference in energy is given by:

$\Delta E = -\left [\frac{2.18 * 10^{-18}}{3^{2}} \right ] - \left [-\frac{2.18 * 10^{-18}}{6^{2}} \right ]$

$\Delta E = 2.18 * 10^{-18}\left [\frac{1}{36}-\frac{1}{9} \right ]$

$\Delta E = 2.18 * 10^{-18}\left [\frac{-1}{12} \right ]$

ΔE = - 0.1816×10^-18 J