Question

three of the strongest lines in the He+ ion spectrum are observed at the following wavelengths: (1) 121.57 nm (2) 164.12 nm (3) 25.64 nm. Find the quantum numbers of the initial and final states for the transitions that give rise to the these three lines. Use equation 4, the wavelengths of lines that can originate from transitions involving any two of the four lowest levels.

Equation 4: delta E = E(upper) - E(lower) = (1.119627 x 10^5 kJ/mole) / wavelength (in nm) or wavelength (in nm) = (1.119627 x 10^5) / delta E (in kJ/mole)

Answer 1

We have,

E_n = - (5248.16/n²) KJ/mol

Therefore,

E₁ = - (5248.16/1²) KJ/mol = -5248.16 KJ/mol

E₂ = - (5248.16/2²) KJ/mol = -1312.04 KJ/mol

E₃ = - (5248.16/3²) KJ/mol = -583.13 KJ/mol

E₄ = - (5248.16/4²) KJ/mol = -328.01 KJ/mol

(1) 121.57 nm

If

n_initial = 4

n_final = 2

Then, E₄-E₂ = -328.01 –(-1312.04) = 984.04 KJ/mol

Now, convert this energy in to wavelength,

= hc/E = [(6.626 x 10^-34) x (2.998 x 10⁸) x (6.023 x 10²³) x (10⁹)]/ (984.04 x 10³ J/mol) = 121.57 nm

Therefore, 121.57 nm corresponding to 4 to 2 transition

(2) 164.12 nm

If

n_initial = 3

n_final = 2

Then, E₃-E₂ = -583.13 –(-1312.04) = 728.91 KJ/mol

Now, convert this energy in to wavelength,

= hc/E = [(6.626 x 10^-34) x (2.998 x 10⁸) x (6.023 x 10²³) x (10⁹)]/ (728.91 x 10³ J/mol) = 164.12 nm

Therefore, 164.12 nm corresponding to 3 to 2 transition

(3) 25.64 nm

If

n_initial = 3

n_final = 1

Then, E₃-E₁ = -583.13 –(-5248.16) = 4665.03 KJ/mol

Now, convert this energy in to wavelength,

= hc/E = [(6.626 x 10^-34) x (2.998 x 10⁸) x (6.023 x 10²³) x (10⁹)]/ (4665.03 x 10³ J/mol) = 25.64 nm

Therefore, 25.64 nm corresponding to 3 to 1 transition