Please help me answer the question below and also show equations and steps of how you came to your answer. Thank you
1. The electron in a hydrogen tom is in the first excited state, when the electron acquires an additional 2.86eV of energy. What is the quantum number n of the state into which the electron moves?
2. A singly ionized helium atom (He+) has only one electron in orbit around the nucleus. What is the radius of the ion when it is in the second exited state?
3. Find the energy (in joules) of the photon that is emitted when the electron in a hydrogen atom undergoes a transition from the n=7 energy level to produce a line in the Paschen series.
Sure, I can help you with those questions. Let's address each question one by one and explain how to get the answer.
1. The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional 2.86 eV of energy. What is the quantum number n of the state into which the electron moves?
To answer this question, we need to use the equation for the energy of an electron in a hydrogen atom:
E = -13.6 eV / n^2
Since the electron acquires an additional 2.86 eV of energy, the new energy can be represented as:
E_new = -13.6 eV / n_new^2
Substituting the given values:
2.86 eV = -13.6 eV / n_new^2
To solve for n_new, we can rearrange the equation:
n_new^2 = -13.6 eV / 2.86 eV
n_new^2 = -13.6 / 2.86
n_new^2 = -4.755
This equation has no real solutions, which means there is no valid quantum number n for the state into which the electron moves. It is not physically possible for the electron to acquire an additional 2.86 eV of energy in the first excited state.
2. A singly ionized helium atom (He+) has only one electron in orbit around the nucleus. What is the radius of the ion when it is in the second excited state?
To find the radius of the ion, we can use the Rydberg formula for hydrogen-like atoms:
1 / λ = R * (Z^2 / n^2 - 1 / m^2)
Where:
- λ is the wavelength of the emitted light in the Paschen series
- R is the Rydberg constant (approximately 1.097 x 10^7 m^-1)
- Z is the atomic number of the nucleus (in this case, Z = 1 for singly ionized helium)
- n is the principal quantum number of the initial energy level
- m is the principal quantum number of the final energy level
In the Paschen series, the transitions start from n = 3 and go to higher energy levels. Since we are interested in the second excited state, we can set n = 3 and m = 4.
Plugging these values into the equation:
1 / λ = R * (1^2 / 3^2 - 1 / 4^2)
Simplifying the equation:
1 / λ = R * (1/9 - 1/16)
1 / λ = R * (7/144)
Solving for λ:
λ = 144 / (7 * R)
Substituting the value of the Rydberg constant, R:
λ = 144 / (7 * 1.097 x 10^7)
Evaluating this expression, we find:
λ ≈ 1.641 x 10^-8 m
The wavelength of the emitted light when the electron undergoes a transition from the n = 3 energy level to the n = 4 energy level is approximately 1.641 x 10^-8 meters.
3. Find the energy (in joules) of the photon that is emitted when the electron in a hydrogen atom undergoes a transition from the n = 7 energy level to produce a line in the Paschen series.
To find the energy of the photon emitted during the transition, we can use the equation:
E = hc / λ
Where:
- E is the energy of the photon
- h is the Planck constant (approximately 6.626 x 10^-34 J*s)
- c is the speed of light (approximately 3.00 x 10^8 m/s)
- λ is the wavelength of the emitted light
To find the wavelength of the emitted light, we can use the formula for the Balmer series. The Balmer series covers transitions from higher energy levels to n = 2. Since we are interested in the Paschen series, we can use the formula for the Balmer limit:
1/λ = R * (1/2^2 - 1/n^2)
Plugging in the given value n = 7:
1/λ = R * (1/2^2 - 1/7^2)
Simplifying the equation:
1/λ = R * (1/4 - 1/49)
1/λ = R * (45/196)
Solving for λ:
λ = 196 / (45 * R)
Substituting the value of the Rydberg constant, R:
λ = 196 / (45 * 1.097 x 10^7)
Evaluating this expression, we find:
λ ≈ 9.288 x 10^-7 m
The wavelength of the emitted light when the electron undergoes a transition from the n = 7 energy level to produce a line in the Paschen series is approximately 9.288 x 10^-7 meters.
Now, let's calculate the energy of the photon using the equation E = hc / λ:
E = (6.626 x 10^-34 J*s) * (3.00 x 10^8 m/s) / (9.288 x 10^-7 m)
E ≈ 2.051 x 10^-19 J
The energy of the photon emitted during the transition is approximately 2.051 x 10^-19 joules.