Photon Energy for the Pfund Series (nf = 5)

Question

a) Find the photon energy and wavelength for the series limit (shortest wavelength) in the Pfund series (nf = 5) for the hydrogen atom. (in eV and nm)

Answer 1

The Pfund series corresponds to electronic transitions in the hydrogen atom where the final energy level \( n_f \) is 5. To find the photon energy and wavelength for the series limit (i.e., the shortest wavelength), we assume the electron is transitioning from \( n_i = \infty \) to \( n_f = 5 \).

### 1. **Photon Energy (in eV)**

The energy difference between two levels in the hydrogen atom is given by the Rydberg formula:

\[

E = 13.6 \, \text{eV} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

\]

For the series limit, \( n_i = \infty \) and \( n_f = 5 \), so the equation simplifies to:

\[

E = 13.6 \, \text{eV} \left( \frac{1}{5^2} - \frac{1}{\infty^2} \right)

\]

\[

E = 13.6 \, \text{eV} \left( \frac{1}{25} \right)

\]

\[

E = 13.6 \, \text{eV} \times 0.04 = 0.544 \, \text{eV}

\]

### 2. **Wavelength (in nm)**

To find the wavelength, use the energy-wavelength relationship:

\[

E = \frac{hc}{\lambda}

\]

Where:

- \( h = 4.1357 \times 10^{-15} \, \text{eV·s} \) (Planck’s constant)

- \( c = 3.0 \times 10^8 \, \text{m/s} \) (speed of light)

- \( \lambda \) is the wavelength in meters

Rearranging the formula for wavelength:

\[

\lambda = \frac{hc}{E}

\]

Substituting the known values:

\[

\lambda = \frac{(4.1357 \times 10^{-15} \, \text{eV·s})(3.0 \times 10^8 \, \text{m/s})}{0.544 \, \text{eV}}

\]

\[

\lambda = \frac{1.2407 \times 10^{-6} \, \text{m·eV}}{0.544 \, \text{eV}}

\]

\[

\lambda = 2.28 \times 10^{-6} \, \text{m} = 2280 \, \text{nm}

\]

### Final Answers:

- **Photon Energy**: 0.544 eV

- **Wavelength**: 2280 nm

This is the photon energy and the wavelength for the series limit (shortest wavelength) in the Pfund series for hydrogen.