Questions: Absorption Emission n=1 to n=3 n=2 to n=1 n=3 to n=5 n=3 to n=2 Answer Bank Ignoring sign, which transition is associated with the greatest energy change? n=3 to n=5 n=1 to n=3 n=2 to n=1 n=3 to n=2

Solution Steps

In atomic physics, the energy change associated with an electron transition between energy levels in an atom is given by the difference in energy between the two levels. The energy levels are quantized and can be calculated using the formula for the energy of an electron in a hydrogen atom:

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

where \( n \) is the principal quantum number of the energy level.

We need to calculate the energy change for each transition listed:

1. \( n=1 \) to \( n=3 \): \[ \Delta E = E_3 - E_1 = \left(-\frac{13.6}{3^2}\right) - \left(-\frac{13.6}{1^2}\right) = -1.5111 + 13.6 = 12.0889 \, \text{eV} \]

2. \( n=2 \) to \( n=1 \): \[ \Delta E = E_1 - E_2 = \left(-\frac{13.6}{1^2}\right) - \left(-\frac{13.6}{2^2}\right) = -13.6 + 3.4 = 10.2 \, \text{eV} \]

3. \( n=3 \) to \( n=5 \): \[ \Delta E = E_5 - E_3 = \left(-\frac{13.6}{5^2}\right) - \left(-\frac{13.6}{3^2}\right) = -0.544 + 1.5111 = 0.9671 \, \text{eV} \]

4. \( n=3 \) to \( n=2 \): \[ \Delta E = E_2 - E_3 = \left(-\frac{13.6}{2^2}\right) - \left(-\frac{13.6}{3^2}\right) = -3.4 + 1.5111 = 1.8889 \, \text{eV} \]

Now, we compare the absolute values of the energy changes calculated:

• \( n=1 \) to \( n=3 \): \( 12.0889 \, \text{eV} \)
• \( n=2 \) to \( n=1 \): \( 10.2 \, \text{eV} \)
• \( n=3 \) to \( n=5 \): \( 0.9671 \, \text{eV} \)
• \( n=3 \) to \( n=2 \): \( 1.8889 \, \text{eV} \)

The greatest energy change is for the transition from \( n=1 \) to \( n=3 \).

Final Answer