The energy levels of one-electron ions are given by the equation:
En=(-2.18*10^(-18)J)(Z^2/n^2)
The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf = 4 state is called the Pickering series, an important series in solar astronomy. Calculate the Pickering series wavelength associated with the excited state ni = 9.
What region of the electromagnetic spectrum does this transition fall in?
radio
microwave
gamma
infrared
UV
visible
X ray
Im suppose to solve for lambda. How do I do that?
After you have delta E, then
dE = hc/wavelength.
gdhs
Alright, let me crunch some numbers and conjure up an answer for you! According to the formula, given that ni = 9 and nf = 4, we can calculate the energy difference between the two states.
ΔE = Ei - Ef = (-2.18 * 10^(-18) J) * [(Z^2 / ni^2) - (Z^2 / nf^2)]
Since we're dealing with a one-electron ion (He+), Z would be 2.
ΔE = (-2.18 * 10^(-18) J) * [(2^2 / 9^2) - (2^2 / 4^2)]
ΔE = (-2.18 * 10^(-18) J) * [(4 / 81) - (4 / 16)]
ΔE = (-2.18 * 10^(-18) J) * [0.0494 - 0.25]
ΔE = (-2.18 * 10^(-18) J) * (-0.2006)
ΔE = 4.36828 * 10^(-19) J
Now, let's find the wavelength using the equation c = λν, where c is the speed of light (3 * 10^8 m/s) and ν is the frequency.
ΔE = hν, where h is Planck's constant (6.626 * 10^(-34) J·s).
ν = ΔE / h = (4.36828 * 10^(-19) J) / (6.626 * 10^(-34) J·s)
ν ≈ 6.605 * 10^14 Hz
Now, to calculate the wavelength:
λ = c / ν = (3 * 10^8 m/s) / (6.605 * 10^14 Hz)
λ ≈ 4.547 * 10^(-7) m
In terms of the electromagnetic spectrum, this wavelength falls in the visible region. So, you'll be able to catch this Pickering series transition with your pretty little eyes! Keep your peepers on the visible spectrum for this one.
To calculate the wavelength of the Pickering series associated with the excited state ni = 9, we can use the formula:
λ = c / ν
where λ is the wavelength, c is the speed of light (approximately 3.00 x 10^8 m/s), and ν is the frequency. The frequency can be calculated using the formula:
ν = ΔE / h
where ΔE is the energy difference between the initial and final states, and h is Planck's constant (approximately 6.63 x 10^-34 J·s).
First, we need to find the energy difference ΔE between the initial state (ni = 9) and the final state (nf = 4). The energy levels of one-electron ions are given by the equation:
En = (-2.18 x 10^(-18) J) × (Z^2 / n^2)
where En is the energy of the nth level, Z is the atomic number (number of protons), and n is the principal quantum number.
Let's substitute the values into the equation:
Ei = (-2.18 x 10^(-18) J) × (1^2 / 9^2) (initial state with ni = 9)
Ef = (-2.18 x 10^(-18) J) × (1^2 / 4^2) (final state with nf = 4)
Now, we can calculate ΔE:
ΔE = Ei - Ef
ΔE = [(-2.18 x 10^(-18) J) × (1^2 / 9^2)] - [(-2.18 x 10^(-18) J) × (1^2 / 4^2)]
Once we obtain the energy difference ΔE, we can calculate the frequency and then the wavelength using the formulas mentioned earlier.
After finding the wavelength, we can determine the region of the electromagnetic spectrum in which this transition falls. The regions from low frequency to high frequency are radio, microwave, infrared, visible, ultraviolet (UV), X-ray, and gamma.
Based on the wavelength obtained, we can determine the corresponding region of the electromagnetic spectrum that the Pickering series transition falls into.
So calculate E for n = 9 (and Z is 2). Call that E1
Then calculate E for n = 4 (and Z is 2). Call that E2.
Subtract the two (disregard the sign), then delta E = hc/wavelength. All of that gives you one line. That's the n = 0 to n = 4 transition.
You will have other lines; for example:
n = 8 to n = 4
n = 7 to n = 4
n = 6 to n = 4
n = 5 to n = 4