Question

Does the hydrogen atom "expand" or "contract" when an electron is excited from the \(n=1\) state to the \(n=3\) state?

Short answer

When an electron is excited from the \(n=1\) state to the \(n=3\) state, the average radius of the electron orbit increases from \(a_0\) to \(9a_0\). Therefore, the hydrogen atom "expands" during this transition.

Step-by-step solution

Write down the formula for the average radius of an electron orbit in a hydrogen atom

The formula for the average radius of an electron orbit in a hydrogen atom is given by: \[ r_n = a_0 n^2 \] where \(r_n\) is the average radius of the orbit for an electron in the principal quantum number (n), \(a_0\) is the Bohr radius (approximately \(5.29 \times 10^{-11}\) meters), and \(n\) is the principal quantum number.

Calculate the average radius of the electron orbit for n=1

Now, we will calculate the average radius of the electron orbit for \(n=1\). Substitute \(n=1\) into the formula: \[ r_1 = a_0 (1)^2 = a_0 \] So, the average radius of the electron orbit for \(n=1\) is equal to the Bohr radius, \(a_0\).

Calculate the average radius of the electron orbit for n=3

Next, we calculate the average radius of the electron orbit for \(n=3\). Substitute \(n=3\) into the formula: \[ r_3 = a_0 (3)^2 = 9a_0 \] The average radius of the electron orbit for \(n=3\) is 9 times the Bohr radius, \(9a_0\).

Compare the radii for n=1 and n=3

Now, we compare the radii for \(n=1\) and \(n=3\): - For \(n=1\), the average radius is \(a_0\). - For \(n=3\), the average radius is \(9a_0\). Since \(9a_0 > a_0\), the average radius of the electron orbit when the electron is excited from the \(n=1\) state to the \(n=3\) state increases.

Conclude whether the hydrogen atom expands or contracts

Because the average radius of the electron orbit increases when the electron is excited from the \(n=1\) state to the \(n=3\) state, we can conclude that the hydrogen atom "expands" during this transition.