The correct answer is option 1: n = 2 to n = 1.
In a hydrogen atom, the electron is in a specific energy level or orbit, denoted by the principal quantum number (n). When the electron transitions from a higher energy level to a lower energy level, it releases energy in the form of electromagnetic radiation.
The energy of the emitted radiation is given by the equation E = (hc) / λ, where E is the energy, h is Planck`s constant, c is the speed of light, and λ is the wavelength of the radiation.
According to this equation, shorter wavelengths correspond to higher energy emissions. Therefore, to find the transition that emits radiation of the shortest wavelength, we need to find the transition with the highest energy difference.
Transition option 1: n = 2 to n = 1 has the highest energy difference compared to the other options provided. This means that it releases the highest amount of energy and emits radiation with the shortest wavelength.
To summarize, the transition of the electron from n = 2 to n = 1 in a hydrogen atom emits radiation of the shortest wavelength, making option 1 the correct answer.