Understanding Electron Transitions in Hydrogen Atom

Electron transitions in a hydrogen atom involve an electron moving from one energy level to another. When an electron moves from a lower energy level to a higher energy level, it absorbs energy. When it moves from a higher energy level to a lower energy level, it releases energy (emits light).

Energy Levels in Hydrogen Atom

The energy levels in a hydrogen atom are quantized and are given by the formula:

\(E_n = -\frac{13.6}{n^2}\) eV

where \(n\) is the principal quantum number (\(n = 1, 2, 3, \dots\)). The ground state is \(n=1\), and the energy levels become less negative (higher energy) as \(n\) increases, approaching 0 eV as \(n \to \infty\).

Calculating Energy Difference

The energy difference (\(\Delta E\)) for a transition from an initial level \(n_i\) to a final level \(n_f\) is given by:

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

For energy to be *required* (absorbed), the final energy level \(n_f\) must be higher than the initial energy level \(n_i\) (\(n_f > n_i\)), resulting in a positive \(\Delta E\). We need to find the transition that gives the largest positive \(\Delta E\).

Analyzing the Given Transitions

Let's calculate the energy difference for each option:

• Option 1: From \(n = \infty\) to \(n = 1\) This is a transition from a very high energy level to the ground state. Since \(n_f < n_i\), energy is released (emission), not required (absorption). \(\Delta E = 13.6 \left(\frac{1}{\infty^2} - \frac{1}{1^2}\right) = 13.6 (0 - 1) = -13.6\) eV. Energy of 13.6 eV is released.
• Option 2: From \(n = 1\) to \(n = 2\) This is a transition from the ground state to the first excited state (\(n_f > n_i\)), so energy is required (absorption). \(\Delta E = 13.6 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6 \left(1 - \frac{1}{4}\right) = 13.6 \left(\frac{3}{4}\right) = 13.6 \times 0.75 = 10.2\) eV. Energy of 10.2 eV is required.
• Option 3: From \(n = 2\) to \(n = 3\) This is a transition from the first excited state to the second excited state (\(n_f > n_i\)), so energy is required (absorption). \(\Delta E = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \left(\frac{9-4}{36}\right) = 13.6 \left(\frac{5}{36}\right) \approx 13.6 \times 0.1389 \approx 1.89\) eV. Energy of approximately 1.89 eV is required.
• Option 4: From \(n = 3\) to \(n = 5\) This is a transition from the second excited state to the fourth excited state (\(n_f > n_i\)), so energy is required (absorption). \(\Delta E = 13.6 \left(\frac{1}{3^2} - \frac{1}{5^2}\right) = 13.6 \left(\frac{1}{9} - \frac{1}{25}\right) = 13.6 \left(\frac{25-9}{225}\right) = 13.6 \left(\frac{16}{225}\right) \approx 13.6 \times 0.0711 \approx 0.97\) eV. Energy of approximately 0.97 eV is required.

Comparing Required Energies

Let's list the required energy for the absorption transitions:

• From \(n=1\) to \(n=2\): 10.2 eV
• From \(n=2\) to \(n=3\): \(\approx\) 1.89 eV
• From \(n=3\) to \(n=5\): \(\approx\) 0.97 eV

Comparing these values, the transition from \(n=1\) to \(n=2\) requires the largest amount of energy (10.2 eV).

The energy difference between consecutive levels decreases as \(n\) increases. The jump from \(n=1\) to \(n=2\) represents the largest energy gap between any two adjacent levels, and also the largest energy required for any single transition starting from the ground state or any excited state to a higher level within the given options.