A hydrogen atom absorbs energy, and an electron is excited to the n=4 shell. When the electron relaxes back to the ground state, n=1, it emits light.




Calculate change in energy of the electron from the excited state back to the ground state and calculate the frequency of the light that was admitted.


To solve this, we can use the Rydberg formula to determine the change in energy when an electron in a hydrogen atom transitions from a higher energy level (excited state) back to the ground state.

### Step 1: Calculate the Change in Energy (\( \Delta E \))
The energy difference between two energy levels in a hydrogen atom is given by:

\[
\Delta E = -13.6 \, \text{eV} \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right)
\]

where:
- \( n_{\text{initial}} = 4 \) (the excited state)
- \( n_{\text{final}} = 1 \) (the ground state)
- \(-13.6 \, \text{eV}\) is the energy of the electron in the ground state of the hydrogen atom.

Substitute the values:

\[
\Delta E = -13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{4^2} \right)
\]

\[
\Delta E = -13.6 \, \text{eV} \left( 1 - \frac{1}{16} \right)
\]

\[
\Delta E = -13.6 \, \text{eV} \times \frac{15}{16}
\]

\[
\Delta E = -12.75 \, \text{eV}
\]

The negative sign indicates energy is released as the electron drops to a lower energy level.

### Step 2: Calculate the Frequency of the Emitted Light
The relationship between the energy of a photon and its frequency is given by:

\[
\Delta E = h \nu
\]

where:
- \( h \) is Planck's constant (\( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)),
- \( \nu \) is the frequency of the emitted light.

First, convert \(\Delta E\) from eV to joules:

\[
\Delta E = -12.75 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = -2.043 \times 10^{-18} \, \text{J}
\]

Now, solve for the frequency \( \nu \):

\[
\nu = \frac{\Delta E}{h}
\]

\[
\nu = \frac{2.043 \times 10^{-18} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}}
\]

\[
\nu = 3.08 \times 10^{15} \, \text{Hz}
\]

### Final Answers:
- **Change in energy**: \( 12.75 \, \text{eV} \)
- **Frequency of emitted light**: \( 3.08 \times 10^{15} \, \text{Hz} \)

Would you like a breakdown of the steps or have any questions about the calculations?

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### Additional Related Questions:
1. What is the wavelength of the emitted photon in this transition?
2. How much energy is absorbed when an electron transitions from \( n=1 \) to \( n=4 \)?
3. What are the energy values for the \( n=2 \) and \( n=3 \) levels in a hydrogen atom?
4. How would the frequency change if the electron only fell to \( n=2 \) before reaching \( n=1 \)?
5. What is the difference in energy between \( n=4 \) to \( n=3 \) compared to \( n=4 \) to \( n=1 \)?

#### Tip:
Remember that higher frequency light corresponds to shorter wavelengths and higher energy.

A hydrogen atom absorbs energy, and an electron is excited to the n=4 shell. When the electron relaxes back to the ground state, n=1, it emits light. Calculate the change in energy of the electron from the excited state back to the ground state and calculate the frequency of the light that was emitted.

Learn how to calculate the energy change and frequency of light emitted as an electron in a hydrogen atom transitions from n=4 to the ground state, n=1. This solution uses the Rydberg formula and covers key concepts in quantum mechanics, suitable for Grades 11-12 or college-level physics.

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