# What is the answer to the following using the Rydberg equation, and why is the answer correct? If the energy of the photons emitted by excited hydrogen atoms is -1.9369 X 10^-18 J, this would correspond most closely to an electron dropping from energy level n2 = ____to n1 = _____. 2 to 1, 3 to 1, 3 to 2, 4 to 2, or 4 to 3 Rydberg equation: (-2.179 X 10^-18)(1/n1^2-1/n2^2)

The Rydberg equation uses a constant and the integers representing two major energy levels of a hydrogen atom to calculate the energy absorbed or emitted when an electron moves between the two energy levels. You're given the energy emitted and need to determine the two integers. You can find the value of the factor (1/n1^2)-(1/n2^2) be dividing the energy emitted by the constant in the equation:

`-1.9369 J = -2.179 x 10^(-18) x (1/n_1^2)-(1/n_2^2)`

`0.8889 = 1/n_1^2-1/n_2^2`

The constant given combines the Rydberg constant, the equation relating frequency to wavelength, and Planck's constant relating the energy of a photon to its frequency.  The units in the equation you gave cancel out to give energy in Joules. For simplicity, I left out the units since they would cancel out to give no units for the energy level integers.

The easiest way to proceed from here it to plug in each pair of energy levels and determine which pair gives a value close to 0.8889. The electron is moving from n2 to n1, closer to the nucleus. In each pair of integers n2 is the first value given and n1 is second.

From 2 to 1: 1-1/4 = 0.75

From 3 to 1: 1-1/9 = 0.8889

From 3 to 2: 1/4 - 1/9 = 0.1389

From 4 to 2: 1/4 - 1/16 =  0.1875

From 4 to 3: 1/9 - 1/16 = 0.0486

The energy level change form 3 to 1 is consistent with the energy emitted.

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