Find the wavelength of radiation required to exite an electron in the ground level of Li2+ to the third energy level
http://www.askiitians.com/iit-jee-structure-of-atom-and-nucleus/solved-examples-of-structure-of-atom/ see problem 1.
To find the wavelength of radiation required to excite an electron in the ground level of Li2+ to the third energy level, we need to calculate the energy difference between these two levels and apply the relationship between energy and wavelength.
Here are the steps to find the wavelength:
1. Determine the energy difference:
The energy difference (ΔE) between two levels can be calculated using the equation:
ΔE = E_final - E_initial
The energy of an electron in a specific energy level can be found using the Rydberg formula:
E = -13.6 eV / n^2
For the ground level of Li2+ (n=1), the initial energy is:
E_initial = -13.6 eV / 1^2 = -13.6 eV
For the third energy level (n=3), the final energy is:
E_final = -13.6 eV / 3^2 = -13.6 eV / 9 = -1.51 eV
Therefore, the energy difference is:
ΔE = -1.51 eV - (-13.6 eV) = 12.09 eV
2. Convert the energy difference to joules:
1 eV = 1.6 × 10^-19 J
So, the energy difference in joules (ΔE_J) is:
ΔE_J = 12.09 eV × (1.6 × 10^-19 J/eV) = 1.9344 × 10^-18 J
3. Calculate the wavelength:
The energy-wavelength relationship is given by the equation:
E = hc / λ, where
E is the energy in joules,
h is the Planck's constant (6.626 × 10^-34 J·s), and
c is the speed of light in a vacuum (3 × 10^8 m/s).
Rearranging the equation to solve for wavelength (λ), we get:
λ = hc / E
Substituting the values, we have:
λ = (6.626 × 10^-34 J·s × 3 × 10^8 m/s) / (1.9344 × 10^-18 J)
λ ≈ 1.02 × 10^-6 m
≈ 1020 nm (nanometers)
Therefore, the wavelength of the radiation required to excite an electron in the ground level of Li2+ to the third energy level is approximately 1020 nm.