To determine the transition corresponding to the emission of the shortest wavelength, we need to consider the energy levels and the corresponding transitions in a hydrogen atom. The energy difference between levels is related to the wavelength of light emitted by:
E = \frac{hc}{\lambda}
Where:
E is the energy difference between two levels.
h is Planck's constant.
c is the speed of light.
\lambda is the wavelength.
The shortest wavelength corresponds to the largest energy difference. In a hydrogen atom, the energy levels are quantized and are calculated by the formula:
E_n = -13.6 \frac{1}{n^2} \, \text{eV}
Transitions to the ground state (n = 1) from higher levels generally have the highest energy difference. Let's look at the transitions depicted:
A: n = 4 \to n = 3
B: n = 4 \to n = 2
C: n = 3 \to n = 2
D: n = 3 \to n = 1
Among these, the transition D: n = 3 \to n = 1 involves the largest drop in energy levels, resulting in the emission of the shortest wavelength of light. Therefore, the correct answer is option D.
