A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.73 nm. It then gives off a photon having a wavelength of 1094 nm. What is the final state of the hydrogen atom?
To determine the final state of a hydrogen atom, we can use the principle of conservation of energy. The energy absorbed by the atom from a photon must be equal to the energy released by the atom when it emits a photon.
Energy of a photon is given by the equation: E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength.
1. For the absorbed photon with a wavelength of 93.73 nm:
- Calculate the energy absorbed: E_absorbed = hc/λ_absorbed.
- Convert the wavelength to meters: λ_absorbed = 93.73 nm × (1 m / 10^9 nm).
- Substitute the values into the equation to find the energy absorbed.
2. For the emitted photon with a wavelength of 1094 nm:
- Calculate the energy released: E_released = hc/λ_released.
- Convert the wavelength to meters: λ_released = 1094 nm × (1 m / 10^9 nm).
- Substitute the values into the equation to find the energy released.
3. Since the energy absorbed must be equal to the energy released, we can set up the equation:
E_absorbed = E_released.
Substitute the values for energy absorbed and released from steps 1 and 2 respectively.
Solving this equation will give us the final state of the hydrogen atom.
To determine the final state of the hydrogen atom, we need to understand the concept of energy levels and transitions.
In the case of hydrogen atoms, electrons occupy various energy levels or orbitals. When an atom absorbs or emits light, it undergoes a transition between these energy levels.
To determine the final state, we first need to identify the initial state of the hydrogen atom. The given information states that the hydrogen atom is in the ground state, which means the electron is initially in the lowest energy level (n=1).
The atom absorbs a photon of light with a wavelength of 93.73 nm. We can use the formula:
E = hc/λ
Where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon.
Substituting the given values:
E = (6.626 x 10^-34 J·s × 2.998 x 10^8 m/s) / (93.73 × 10^-9 m)
E ≈ 2.089 x 10^-18 J
This energy corresponds to the energy difference between the ground state and an excited state of the hydrogen atom.
Next, we are told that the hydrogen atom gives off a photon with a wavelength of 1094 nm. Using the same formula, we can calculate the energy of this emitted photon:
E = (6.626 x 10^-34 J·s × 2.998 x 10^8 m/s) / (1094 × 10^-9 m)
E ≈ 1.815 x 10^-18 J
Now, to determine the final state of the hydrogen atom, we need to find the energy level that corresponds to an energy difference of 1.815 x 10^-18 J relative to the ground state energy.
Using the formula for energy levels of hydrogen atoms:
E = -13.6 eV / n^2
Where E is the energy of the energy level, -13.6 eV is the ionization energy of a hydrogen atom, and n is the principal quantum number.
Let's solve for n:
1.815 x 10^-18 J = -13.6 eV / n^2
Converting eV to Joules (1 eV = 1.602 x 10^-19 J):
1.815 x 10^-18 J = -13.6 × 1.602 x 10^-19 J / n^2
n^2 ≈ -13.6 × 1.602 x 10^-19 J / 1.815 x 10^-18 J
n^2 ≈ 0.119
Taking the square root:
n ≈ √0.119
n ≈ 0.345
Since the principal quantum number (n) cannot be a fraction or negative, we take the next highest whole number. Therefore, the final state of the hydrogen atom is n = 1.
Hence, the final state of the hydrogen atom remains in the ground state (n = 1) after absorbing and emitting the given photons.
w = wavelength
1/w = 1.0973E7(1/1 - 1/x^2)
Convert 93.73 nm to m and solve for x. That will tell you to which orbit the electron has been moved.
Then, 1/w = 1.0973(1/y^2 - 1/x^2).
Solve for y.