Consider the Li2+ ion.
(a) Calculate the wavelength of light emitted when the electron jumps from the n=4 level
to the n=2 level.
(b) What is the ionization energy of Li2+ ?
To calculate the wavelength of light emitted when an electron jumps from one energy level to another, we can use the equation:
ΔE = Rh (1/n1² - 1/n2²)
where ΔE is the energy change, Rh is the Rydberg constant (2.18 x 10^-18 J), and n1 and n2 are the initial and final energy levels, respectively.
(a) Let's calculate the energy change for the electron jumping from the n=4 level to the n=2 level:
ΔE = Rh (1/2² - 1/4²)
ΔE = Rh (1/4 - 1/16)
ΔE = Rh (3/16)
Now, we can calculate the wavelength of the light emitted using the equation:
λ = c / ν
where λ is the wavelength, c is the speed of light (3.00 x 10^8 m/s), and ν is the frequency of the light emitted.
Since we have the energy change (ΔE), we can relate it to the frequency using the equation:
ΔE = hν
where h is Planck's constant (6.63 x 10^-34 J·s).
Therefore, we can rewrite the equation for the wavelength:
λ = c / (ΔE / h)
Now, we can substitute the values and calculate the wavelength:
λ = (3.00 x 10^8 m/s) / ((Rh (3/16)) / (6.63 x 10^-34 J·s))
Simplifying the expression:
λ = (3.00 x 10^8 m/s) / ((2.18 x 10^-18 J) * (3/16) / (6.63 x 10^-34 J·s))
Calculating the expression:
λ ≈ 1.21 x 10^-7 m
Therefore, the wavelength of light emitted when the electron jumps from the n=4 level to the n=2 level in Li2+ is approximately 1.21 x 10^-7 meters.
(b) The ionization energy is the energy required to remove an electron from an ion. In this case, we are dealing with the Li2+ ion, which means it already has a positive charge of +2.
The ionization energy of Li2+ is the energy required to remove an electron from the Li2+ ion and turn it into a neutral Li atom. We can calculate it using the energy formula:
Ionization energy = - ΔE
where ΔE is the energy change when an electron is removed from the Li2+ ion.
In this case, since we are removing an electron, the energy change will be positive. We can use the same formula we used in part (a) to calculate ΔE:
ΔE = Rh (1/n1² - 1/n2²)
In this case, since we want to remove the electron completely, n2 will be infinity (∞).
ΔE = Rh (1/n1² - 1/∞²)
ΔE = Rh (1/n1²)
Substituting the values:
ΔE = (2.18 x 10^-18 J) (1/4²)
ΔE = (2.18 x 10^-18 J) (1/16)
Calculating the expression:
ΔE ≈ 1.36 x 10^-19 J
Therefore, the ionization energy of Li2+ is approximately 1.36 x 10^-19 Joules.