Atoms of ionized lithium gas (Li2+) are struck by neutrons moving at a velocity of 1.46e5 m/s. Calculate the shortest wavelength in the emission spectrum of Li2+ under these circumstances. You can assume that all electrons start in the ground state.
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To calculate the shortest wavelength in the emission spectrum of Li2+ under these circumstances, we need to consider the transition of electrons from higher energy levels to lower energy levels.
The energy change in the transition of an electron between two energy levels is given by the equation:
ΔE = Ef - Ei
Where ΔE is the energy change, Ef is the final energy level, and Ei is the initial energy level.
The energy of a photon can be related to its wavelength using the equation:
E = hc/λ
Where E is the energy, h is the 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.
To find the shortest wavelength, we need to find the transition that results in the highest energy photon. Since all electrons start in the ground state, we need to find the highest energy level in Li2+.
Li2+ has two electrons, so we need to consider the energy levels for both electrons. For a hydrogen-like ion, the energy of an electron in an energy level n is given by:
En = -(13.6 eV) Z^2/n^2
Where En is the energy of the electron, Z is the atomic number of the nucleus (in this case, 3 for lithium), and n is the principal quantum number.
The ground state for a hydrogen-like ion is when n = 1. Therefore, we need to calculate the energy of the first excited state (n = 2) for both electrons.
For Li2+, the energy of the first excited state for each electron is:
E2 = -(13.6 eV) (3^2/2^2) = -34 eV
Next, we need to calculate the energy change (ΔE) for an electron transitioning from the first excited state (n = 2) to the ground state (n = 1):
ΔE = E1 - E2
= 0 - (-34 eV)
= 34 eV
Finally, we can calculate the shortest wavelength (λ) using the energy change (ΔE):
E = hc/λ
34 eV = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s)/λ
Solving for λ gives:
λ = ((6.626 x 10^-34 J·s)(3.00 x 10^8 m/s))/(34 eV)
Converting eV to Joules (1 eV = 1.6 x 10^-19 J) and evaluating:
λ = ((6.626 x 10^-34 J·s)(3.00 x 10^8 m/s))/(34 eV)(1.6 x 10^-19 J/eV)
= 1.23 x 10^-7 m
Therefore, the shortest wavelength in the emission spectrum of Li2+ under these circumstances is approximately 1.23 x 10^-7 m.