1)What is the longest wavelength light capable of ionizing a hydrogen atom in the n = 6 state
2)a. Calculate the energy of the electron in the ground state of single ionized helium, which has an atomic number Z = 2.
b. Calculate the radius of its ground-state orbit
1) Solve h c/L = (Rydberg)/6^2 for L
The Rydberg energy is equivalent to 13.6 eV, but you will have to convert it to Joules
2) a) The ground state energy of He+ is 2^2 = 4 times below that of hydrogen in the n=1 quantum level.
b) half the radius of the n=1 Bohr orbit of hydrogen
I thank you immensely for your help.
what is h and c?
1) To determine the longest wavelength light capable of ionizing a hydrogen atom in the n = 6 state, we need to find the energy difference between the n = 6 and n = ∞ states, and then convert it to a wavelength.
The energy difference between any two states in the hydrogen atom can be calculated using the formula:
ΔE = (13.6 eV) * (1/n₁² - 1/n₂²)
Where ΔE is the energy difference in electron volts (eV), and n₁ and n₂ are the principal quantum numbers representing the initial and final states, respectively.
For this question, we are interested in the energy difference between n = 6 and n = ∞. So, plugging in the values:
ΔE = (13.6 eV) * (1/6² - 1/∞²)
= (13.6 eV) * (1/36 - 0)
= (13.6 eV) * (1/36)
≈ 0.378 eV
Now, let's convert this energy to wavelength. We can use the equation:
E = h * c / 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 (3.0 x 10⁸ m/s), and λ is the wavelength.
Converting the energy from electron volts to joules:
1 eV = 1.602 x 10⁻¹⁹ J
0.378 eV ≈ 0.378 * (1.602 x 10⁻¹⁹ J)
≈ 6.06 x 10⁻²⁰ J
Rearranging the equation to solve for λ:
λ = h * c / E
Plugging in the values:
λ = (6.626 x 10⁻³⁴ J.s) * (3.0 x 10⁸ m/s) / (6.06 x 10⁻²⁰ J)
≈ 3.31 x 10¹⁷ meters
Therefore, the longest wavelength light capable of ionizing a hydrogen atom in the n = 6 state is approximately 3.31 x 10¹⁷ meters.
2a) To calculate the energy of the electron in the ground state of single ionized helium (He²⁺), we can use the formula:
E = -13.6 eV / (Z²)
Where Z is the atomic number of the helium atom (Z = 2 in this case).
Plugging in the values:
E = -13.6 eV / (2²)
= -13.6 eV / 4
= -3.4 eV
Therefore, the energy of the electron in the ground state of single ionized helium is -3.4 eV.
2b) To calculate the radius of the ground-state orbit, we can use the Bohr radius formula:
r = (5.29 x 10⁻¹¹ meters) * n² / Z
Where r is the radius of the orbit, n is the principal quantum number (1 in the ground state), and Z is the atomic number of the helium atom (Z = 2).
Plugging in the values:
r = (5.29 x 10⁻¹¹ meters) * (1²) / 2
= (5.29 x 10⁻¹¹ meters) / 2
= 2.645 x 10⁻¹¹ meters
Therefore, the radius of the ground-state orbit in single ionized helium is approximately 2.645 x 10⁻¹¹ meters.