The longest and shortests wavelengths of the Lyman series of singly ionized Helium
http://en.wikipedia.org/wiki/Lyman_series
Ignore my post, Miss X, I didn't read the question. Duh.
Find the wave length of the first line of the series that corresponds to the Lyman series in a single ionized helium atom where the helium nucelus is four times of the hydrogen nucelus and the change of the helium nucelus is four times of the hydrogen nucelus
To find the longest and shortest wavelengths of the Lyman series of singly ionized Helium, we need to understand the concept of the Lyman series and its relationship with the energy levels of an atom.
The Lyman series is a series of spectral lines in the ultraviolet region of the electromagnetic spectrum. It is specifically associated with the transitions of electrons in hydrogen-like atoms, where only one electron is present.
The energy levels for a hydrogen-like atom, including singly ionized Helium, can be represented by the equation:
E = -13.6 eV/n^2
where E is the energy of the level, -13.6 eV is the ionization energy of hydrogen, and n represents the principal quantum number of the energy level.
The Lyman series corresponds to the transitions where the electron jumps from a higher energy level to the n = 1 energy level. The energy difference between these levels is given by:
ΔE = E_final - E_initial
For the Lyman series in singly ionized Helium, the initial energy level (E_initial) can be any higher energy level (n > 1). The final energy level (E_final) is always the n = 1 level.
To find the longest wavelength, we need to find the transition with the smallest energy difference (ΔE). The smallest energy difference corresponds to the largest transition, resulting in the longest wavelength. This occurs when n is the largest possible value.
In this case, the largest value of n is 2 since we are considering singly ionized Helium. Plugging this value into the energy equation, we can calculate the energy difference:
ΔE = E_final - E_initial
ΔE = (-13.6 eV/1^2) - (-13.6 eV/2^2)
ΔE = -13.6 eV - (-3.4 eV)
ΔE = -10.2 eV
To find the corresponding wavelength, we can use the formula:
λ = hc/ΔE
where λ is the wavelength, 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 ΔE is the energy difference in Joules.
Plugging in the values, we can calculate the longest wavelength:
λ = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (10.2 eV * 1.6 x 10^-19 J/eV)
λ = 1.24 x 10^-7 m
Therefore, the longest wavelength for the Lyman series of singly ionized Helium is approximately 124 nm.
To find the shortest wavelength, we need to find the transition with the largest energy difference (ΔE), resulting in the smallest transition and shortest wavelength. This occurs when the initial energy level (E_initial) is the smallest value, which is n = 2.
Again, plugging this value into the energy equation, we can calculate the energy difference:
ΔE = E_final - E_initial
ΔE = (-13.6 eV/1^2) - (-13.6 eV/2^2)
ΔE = -13.6 eV - (-3.4 eV)
ΔE = -10.2 eV
Using the same wavelength formula as before, we can calculate the shortest wavelength:
λ = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (10.2 eV * 1.6 x 10^-19 J/eV)
λ = 1.24 x 10^-8 m
Therefore, the shortest wavelength for the Lyman series of singly ionized Helium is approximately 12.4 nm.
1/4 of the corresponding wavelengths for hydrogen. The shortest wavelength will be the series limit.
The Lyman alpha line of H is at 121.6 nm