For the emission spectrum of Be3+, calculate the lowest wavenumber νˉ (in inverse meters) of light produced by electron transitions between n=2, n=3, and n=4.
w = wavelength
1/w = RZ^2(1/n1^2 - 1/n2^2)
R = 1.09737E7
Z = 4
n1^2= 3^2= 9
n2^2 = 4^2 = 16
8.56*10^6
amar chauhan
What are the concepts used in these?
To calculate the lowest wavenumber (νˉ) of light produced by electron transitions between energy levels in Be3+, we can use the Rydberg formula:
1/νˉ = R_H [Z^2 / (n_1^2) - Z^2 / (n_2^2)]
where νˉ is the wavenumber in inverse meters, R_H is the Rydberg constant (approximately 109677 inverse meters), Z is the atomic number of the element (in this case Z=4 for Be), and n_1 and n_2 are the initial and final energy levels, respectively.
For the lowest wavenumber, we need to find the largest difference in energy levels available for Be3+. In this case, the largest difference occurs when the electron jumps from the highest energy level (n=4) to the lowest energy level (n=2).
Substituting n_1 = 4 and n_2 = 2 into the Rydberg formula, we get:
1/νˉ = R_H [4^2 / (2^2) - 4^2 / (4^2)] = R_H [16/4 - 16/16] = R_H (4 - 1) = 3R_H
Now we can substitute the value for the Rydberg constant into the equation:
1/νˉ = 109677 * 3 = 329031 inverse meters
Therefore, the lowest wavenumber (νˉ) of light produced by electron transitions between n=2, n=3, and n=4 in Be3+ is approximately 329031 inverse meters.