The Paschen series corresponds to electron transitions to the $n = 3$ energy level in a hydrogen atom. For this problem, you are asking for the energy difference for a transition from $n = 6$ to $n = 3$.

The energy of an electron in the hydrogen atom is given by the formula:

$E_n = -\frac{13.6 \, \text{eV}}{n^2}$

where $n$ is the principal quantum number, and the energy is in electron volts (eV).

To find the energy difference between $n = 6$ and $n = 3$, calculate the energy of both levels and then subtract:

$\Delta E = E_6 - E_3$

Step 1: Calculate $E_6$

$E_6 = -\frac{13.6 \, \text{eV}}{6^2} = -\frac{13.6 \, \text{eV}}{36} = -0.3778 \, \text{eV}$

Step 2: Calculate $E_3$

$E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.5111 \, \text{eV}$

Step 3: Calculate the energy difference

$\Delta E = E_3 - E_6 = (-1.5111 \, \text{eV}) - (-0.3778 \, \text{eV})$ $\Delta E = -1.5111 + 0.3778 = -1.1333 \, \text{eV}$

Thus, the energy difference for the transition from $n = 6$ to $n = 3$ is:

$\boxed{1.133 \, \text{eV}}$

Do you want more details or have any questions?

Here are 5 related questions:

1. How does the Paschen series differ from the Lyman or Balmer series?
2. What are the units of energy commonly used in atomic transitions?
3. How does the energy level spacing change as $n$ increases in hydrogen?
4. What is the physical meaning of a negative energy value in this context?
5. How does the wavelength of light emitted relate to the energy difference?

Tip: The greater the difference in energy levels, the shorter the wavelength of the emitted photon!