How much energy is required to raise the hydrogen atom from the ground state \(n=1\) to the first excited state \(n=2\) ? What is the wavelength of the line emitted if the atom returned back to the ground state.

The Rydberg formula is essential for understanding the energy levels of a hydrogen atom. It provides a way to calculate the energy associated with an electron in a specific energy level or orbit, identified by the principal quantum number, denoted as \( n \). This formula is given by:
\[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]
Here, 13.6 eV represents the ionization energy of hydrogen, which is the energy required to completely remove the electron from the hydrogen atom when it is in the ground state (\( n = 1 \)).
The negative sign indicates that the energy is bound or less than zero; a bound electron within an atom has lower potential energy compared to a free electron.

Let’s break it down further:

• Principal Quantum Number (\( n \)): This integer value represents the different energy levels or orbits an electron can occupy in a hydrogen atom.
• Energy Levels (\( E_n \)): These are the discrete energies electrons can have; they aren’t continuous, meaning electrons can only exist in specific orbits corresponding to specific energies.

Using the Rydberg formula, you can find the energy at any energy level \( n \). This is crucial when studying transitions between different energy levels, like moving from the ground state to an excited state or vice versa.

Energy levels in a hydrogen atom are quantized, meaning electrons can only occupy certain energy states. To determine the required energy to move an electron between these levels, such as from the ground state (\( n = 1 \)) to the first excited state (\( n = 2 \)), we calculate the difference in their energies.

First, calculate the energy at the ground state using the Rydberg formula:
\[ E_1 = -\frac{13.6 \text{ eV}}{1^2} = -13.6 \text{ eV} \]
Next, calculate the energy at the first excited state:
\[ E_2 = -\frac{13.6 \text{ eV}}{2^2} = -\frac{13.6 \text{ eV}}{4} = -3.4 \text{ eV} \]
To find the energy needed to raise the hydrogen atom from \( n = 1 \) to \( n = 2 \), determine the energy difference:
\[ \text{Energy Difference} = E_2 - E_1 = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 10.2 \text{ eV} \]
This energy difference of 10.2 eV is what's required for the transition from the ground state to the first excited state.

Key points to remember about energy levels:

• The energy level increases as \( n \) increases, meaning electrons in higher orbits have higher energy.
• The difference in energy levels (the energy gap) determines the energy required for transitions or the energy emitted/absorbed when such transitions occur.

After an electron transitions back from an excited state to a lower energy level, such as from \( n = 2 \) back to \( n = 1 \), it emits energy in the form of a photon. The energy of this photon corresponds to the energy difference between the two states, determined by:

• Energy of the photon (\( E \)) = 10.2 eV (found earlier)

To find the wavelength (\( \lambda \)) of this emitted photon, we use the relation:
\[ E = \frac{hc}{\lambda} \]
Rearrange it to solve for \( \lambda \):
\[ \lambda = \frac{hc}{E} \]
where:

• \( h \) is Planck's constant, approximately \( 4.135667696 \times 10^{-15} \text{ eV·s} \)
• \( c \) is the speed of light, about \( 3 \times 10^8 \text{ m/s} \)
• \( E \) is the energy difference, 10.2 eV

Substituting the values in, we get:
\[ \lambda = \frac{4.135667696 \times 10^{-15} \text{ eV·s} \times 3 \times 10^8 \text{ m/s}}{10.2 \text{ eV}} \approx 1.22 \times 10^{-7} \text{ m} \]
This means the wavelength of the emitted photon is around 122 nm.

Key takeaways:

• The wavelength of light emitted or absorbed during transitions provides information about the underlying energy levels in an atom.
• Different wavelengths correspond to different colors in the spectrum, with shorter wavelengths indicating higher energy transitions.