A. Define the following terms: i. Stationary state, ii. Ground state, iii. Excitation energy. b. The energy levels in a hydrogen atom are given by the equation: E_n = rac−13.6n2 eV. If the electron mo

Here are the solutions to the questions:

12. a. Define the following terms:

• i. Stationary state: A state in which an electron in an atom has a definite, constant energy and does not radiate energy.
• ii. Ground state: The lowest possible energy state that an electron can occupy in an atom.
• iii. Excitation energy: The minimum energy required to move an electron from its ground state to a higher energy level (an excited state).

12. b. The energy levels in a hydrogen atom are given by the equation: $E_n = \frac{-13.6}{n^2}$ eV. If the electron moves from the third excited state to the ground state, calculate:
Given:
$E_n = \frac{-13.6}{n^2}$ eV
Electron charge, $e = 1.6 \times 10^{-19}$ C
Speed of light, $c = 3.0 \times 10^8$ m/s
Planck's constant, $h = 6.6 \times 10^{-34}$ Js

The ground state corresponds to $n=1$.
The first excited state corresponds to $n=2$.
The second excited state corresponds to $n=3$.
The third excited state corresponds to $n=4$.

The electron moves from $n_i = 4$ (third excited state) to $n_f = 1$ (ground state).

• i. The energy of the radiation emitted ($E_{photon}$):

  Step 1: Calculate the energy of the initial state ($n=4$).

  $$E_4 = \frac{-13.6}{4^2} = \frac{-13.6}{16} = -0.85 \text{ eV}$$

  Step 2: Calculate the energy of the final state ($n=1$).

  $$E_1 = \frac{-13.6}{1^2} = \frac{-13.6}{1} = -13.6 \text{ eV}$$

  Step 3: Calculate the energy of the emitted photon.

  $$E_{photon} = E_i - E_f = E_4 - E_1$$ $$E_{photon} = -0.85 \text{ eV} - (-13.6 \text{ eV}) = -0.85 + 13.6 = 12.75 \text{ eV}$$

  Step 4: Convert the energy from electron volts (eV) to Joules (J).

  $$E_{photon} = 12.75 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV})$$ $$E_{photon} = \boxed{\text{2.04} \times 10^{-18} \text{ J}}$$

• ii. The wavelength of the radiation ($\lambda$):

  Step 1: Use the formula relating energy, Planck's constant, speed of light, and wavelength: $E = \frac{hc}{\lambda}$.
  Rearrange to solve for $\lambda$: $\lambda = \frac{hc}{E_{photon}}$.

  Step 2: Substitute the values.

  $$$$