Question

Assume that a hydrogen atom's electron has been excited to the \(n=6\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

Short answer

In this problem, a hydrogen atom's electron is excited to the \(n=6\) energy level. The electron can de-excite to any lower energy level (1, 2, 3, 4, or 5) and emit light with different wavelengths. For an atom with an initial energy level \(n\), there are \((n-1)\) possible transitions. In our case, there are \((6-1)=5\) different wavelengths of light that can be emitted as the excited atom loses energy.

Step-by-step solution

Determine the initial energy level (n_initial)

The electron is excited from its ground state to n=6 energy level. So, the initial energy level (n_initial) is 6.

Determine the possible final energy levels (n_final)

As the atom de-excites, it will lose energy in the form of light emission. The electron can move from n_initial = 6 to any lower energy level (n_final). Thus, the possible final energy levels are 1, 2, 3, 4, and 5.

Calculate the number of possible transitions

To find the number of possible transitions, we need to find the number of combinations that can be formed by choosing one final level from the list of possible final levels. There are 5 possible final levels for the electron (1, 2, 3, 4, and 5), so one could have 5 possible transitions. For an atom with initial energy level n, there are \((n-1)\) possible transitions to emit light. In our case, the number of transitions is \((6-1) = 5\). So, there are 5 different wavelengths of light that can be emitted as the excited atom loses energy.