Question

1.817 x 10-22 kJ of energy is released for an electron in a hydrogen atom to make a transition from a higher energy level to the n=3 energy level. What energy level did the electron in the atom of hydrogen originate before the transition?

I know that the answer is n=6 based on an answer key, but I do not know how to reach this conclusion.

Answer 1

The electron in a hydrogen atom originated from the
`\(n = 3\)` energy level before transitioning. The energy released,
`\(1.817 * 10^(-22) \, \text{kJ}\),` corresponds to a shift to
`\(n = 3\).`

The energy of a photon emitted or absorbed during a transition in a hydrogen atom can be calculated using the Rydberg formula:

`\[ \Delta E = - R_H \left( (1)/(n_f^2) - (1)/(n_i^2) \right) \]`

where:

-
`\(\Delta E\)` is the energy of the photon,

`\(R_H\)` is the Rydberg constant for hydrogen
`(\(2.179 * 10^(-18) \, \text{J}\)),`

-
`\(n_f\)` is the final energy level,

-
`\(n_i\)` is the initial energy level.

In this case, we're given
`\(\Delta E = 1.817 * 10^(-22) \, \text{J}\)` and
`\(n_f = 3\).` We need to find
`\(n_i\).`Rearranging the formula to solve for
`\(n_i\)`, we get:

`\[ (1)/(n_i^2) = (1)/(n_f^2) - (\Delta E)/(R_H) \]`

Substituting the given values:

`\[ (1)/(n_i^2) = (1)/(3^2) - (1.817 * 10^(-22))/(2.179 * 10^(-18)) \]`

Now, solve for
`\(n_i\)`

`\[ (1)/(n_i^2) = (1)/(9) - (1.817 * 10^(-22))/(2.179 * 10^(-18)) \]`

`\[ (1)/(n_i^2) = (1)/(9) - 8.36 * 10^(-5) \]`

`\[ (1)/(n_i^2) = 0.111 - 8.36 * 10^(-5) \]`

`\[ (1)/(n_i^2) \approx 0.1109164 \]`

`\[ n_i^2 \approx (1)/(0.1109164) \]`

`\[ n_i^2 \approx 9.028 \]`

`\[ n_i \approx √(9.028) \]`

`\[ n_i \approx 3.00 \]`

So, the electron originated from the \(n = 3\) energy level before the transition.