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.
Those are three questions, not one.
1. The first excited state of hydrogen is the n=2 level and that has an energy of -13.6/2^2 = -3.40 eV. Adding 2.86 eV to that results in an energy of -0.54 eV
The quantum level n' of that state is given by
-13.6/n'^2 = -0.54
Solve for n' and round off to the nearest integer. Orbital quantum numbers must be integers.
2. Look for the equation that tells you how the "Bohr" radius of a hydrogenic (one-electron, Rydberg) atom depends upon the nuclear atomic number, Z (which in this case, is 2) and the principal quantum number (in this case, also 2).
http://en.wikipedia.org/wiki/Rydberg_atom
has the formula you need.
H9int: For the Paschen series, the quantum number of the lower state is n = 3. The upper state of the line in question is n=7. For the photon energy, use the formula
E = 13.6 eV * (1/3^2 - 1/7^2)
Sure, I can help you with these questions! Let's go step by step:
1. To find the quantum number (n) of the state into which the electron moves, we can use the equation for the energy of an electron in the hydrogen atom:
E = -13.6/n^2
We are given that the electron acquires an additional 2.86eV of energy. We need to convert this energy into Joules since the equation requires energy in SI units.
1eV = 1.6 x 10^-19 J
So, 2.86eV = 2.86 x 1.6 x 10^-19 J = 4.576 x 10^-19 J
We can set the final energy equal to the initial energy plus the additional energy:
E_final = E_initial + additional energy
Using the equation for energy in the hydrogen atom, we can rewrite this as:
-13.6/n_final^2 = -13.6/n_initial^2 + 4.576 x 10^-19 J
Since the electron is initially in the first excited state (n=2), we can substitute n_initial = 2:
-13.6/n_final^2 = -13.6/2^2 + 4.576 x 10^-19 J
Simplifying further:
-13.6/n_final^2 = -13.6/4 + 4.576 x 10^-19 J
Now we can solve for n_final. Rearranging the equation:
1/n_final^2 = 1/4 - (4.576 x 10^-19 J)/13.6
n_final^2 = 4/(1/4 - (4.576 x 10^-19 J)/13.6)
n_final^2 = 4/(0.25 - (4.576 x 10^-19 J)/13.6)
n_final^2 = 16/(0.25 - (4.576 x 10^-19 J)/13.6)
n_final^2 = 16/(0.25 - (4.576 x 10^-19 J)/13.6)
n_final^2 = 16/(0.25 - (4.576 x 10^-19 J)/13.6)
n_final^2 ≈ 6
Therefore, the quantum number of the state into which the electron moves is n = 6.
2. To find the radius of the singly ionized helium ion (He+) when it is in the second excited state, we can use the equation for the radius of the Bohr orbit:
r = 0.529 Å * n^2/Z
where r is the radius, n is the principal quantum number, and Z is the atomic number.
For a singly ionized helium atom (He+), Z = 2 (the atomic number of helium).
So, for the second excited state (n=3):
r = 0.529 Å * 3^2/2
r = 0.529 Å * 9/2
r = 2.6465 Å
Therefore, the radius of the He+ ion when it is in the second excited state is approximately 2.6465 Å.
3. To find the energy of the photon 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, we can use the equation:
ΔE = E_final - E_initial = -13.6 (1/n_final^2 - 1/n_initial^2)
We are given that the initial energy level is n = 7, so n_initial = 7.
Using n_final = ∞ for the Paschen series (since the electron goes to a higher energy level):
ΔE = -13.6 (1/∞^2 - 1/7^2)
ΔE = -13.6 (0 - 1/49)
Simplifying:
ΔE = -13.6/49
To convert this energy into joules, we multiply by -1 (to get a positive value) and the electron charge magnitude (1.6 x 10^-19 C):
ΔE = (13.6/49) * 1.6 x 10^-19 J/C
ΔE ≈ 4.67 x 10^-19 J
Therefore, the energy of the photon emitted when the electron in a hydrogen atom undergoes this transition is approximately 4.67 x 10^-19 J.
Let me know if there is anything else I can help you with!
Sure! Let's go through each question one by one and explain how to find the answers:
1. The energy of an electron in the excited states of a hydrogen atom can be calculated using the formula:
E = -13.6/n^2
where E is the energy in electron volts (eV), and n is the principal quantum number. We can set up an equation to find the value of n for the excited state:
E_initial + 2.86 eV = -13.6/n^2
Solving this equation for n, we can find the value of n for the state into which the electron moves.
2. The radius of the orbit of an electron in a hydrogen-like atom (such as He+) can be calculated using the formula:
r = 0.529 * n^2 / Z
where r is the radius in Ångstroms (Å), n is the principal quantum number, and Z is the atomic number. Since the He+ ion has only one electron, the atomic number Z is 2. To find the radius when the ion is in the second excited state, we substitute n = 2 into the equation.
3. The energy of a photon can be calculated using the formula:
E = h * c / λ
where E is the energy in joules (J), 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 of the photon in meters (m). The transition in the Paschen series from n = 7 can be represented by n = 7 to n = 3.
First, we find the energy difference between these two levels using the formula from question 1:
ΔE = Ef - Ei = -13.6 * (1/3^2 - 1/7^2)
Then, we can use the energy difference to calculate the energy of the photon emitted using the equation:
E = ΔE * 1.6 x 10^-19 J
This will give us the energy of the photon in joules.
Remember to always keep track of the units and constants used in the equations to ensure accurate calculations.