Electrons of energies \(10.2 \mathrm{eV}\) and \(12.09 \mathrm{eV}\) can \text { cause radiation to be emitted from hydrogen atoms. Calculate in each case, the principal quantum number of the orbit to whichan electron in the hydrogen atom is raised and the wavelength of radiation if it drops back to the ground state.

The energy levels of a hydrogen atom are quantized, meaning the electrons can only occupy certain energy levels or orbits. These levels are defined by integers known as the principal quantum numbers, denoted as \( n \). The energy associated with each level can be calculated using the formula: \[ E_n = -13.6 \frac{1}{n^2} \text{ eV} \] where \( E_n \) is the energy at the principal quantum number \( n \). The negative sign represents the fact that the electron is bound to the nucleus. The closer the electron is to the nucleus (lower \( n \)), the more negative and hence lower is the energy. When an electron absorbs energy, it moves to a higher energy level (larger \( n \)). Conversely, when it falls back to a lower energy level, it emits energy usually in the form of radiation. Understanding these energy levels helps to calculate various properties and transitions of electrons within the atom.

The principal quantum number, represented by \( n \), is crucial in determining the energy level of an electron in a hydrogen atom. It is a positive integer (1, 2, 3, ...) that signifies the orbit in which an electron resides. The larger the quantum number, the higher the energy level and the farther the electron is from the nucleus. For example:

• When \( n = 1 \), the electron is in the first energy level (ground state).
• When \( n = 2 \), it is in the second energy level.
• When \( n = 3 \), it is in the third energy level, and so on.

To find the principal quantum number of an electron given its energy, we rearrange the energy level formula: \[ E_n = -13.6 \frac{1}{n^2} \text{ eV} \] Given the energy, solve for \( n \) by isolating it on one side of the equation and taking the square root. For an energy of 10.2 eV, solving the equation: \[ -13.6 \left( \frac{1}{n^2} \right) + 13.6 = 10.2 \] yields \( n = 2 \).

When an electron falls from a higher energy level to a lower one, it emits energy in the form of radiation with a specific wavelength. This can be calculated using the Rydberg formula: \[ \frac{1}{\text{wavelength}} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H \) is the Rydberg constant (\( 1.097 \times 10^7 \text{m}^{-1} \)), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level. For example, if an electron falls from \( n = 2 \) to \( n = 1 \), substitute these values: \[ \frac{1}{\text{wavelength}} = 1.097 \times 10^7 \left( 1 - \frac{1}{4} \right) \] Solving this gives the wavelength: \[ \text{wavelength} = 1.216 \times 10^{-7} \text{ m} \text{ or } 121.6 \text{ nm} \] This formula helps to determine the exact wavelength of the radiation emitted during an electron transition within a hydrogen atom.