one of the emission spectra lines for Be+3 has a wavelength of 253.4 nm for an electronic transition that begins in the state with n=5. what is the principal quantum number of the lower energy state corresponding to this emission? round your answer to the nearest whole number
To find the principal quantum number of the lower energy state corresponding to the emission line with a wavelength of 253.4 nm for Be+3, we can use the Rydberg formula:
1/λ = R (1/n1^2 - 1/n2^2)
Where:
- λ is the wavelength of the emission line (253.4 nm = 253.4 x 10^-9 m)
- R is the Rydberg constant (1.097 x 10^7 m^-1)
- n1 is the principal quantum number of the lower energy state (unknown)
- n2 is the principal quantum number of the higher energy state (n2 = 5)
Rearranging the formula to solve for n1:
1/n1^2 = 1/(λ*R) + 1/n2^2
1/n1^2 = 1/(253.4 x 10^-9 * 1.097 x 10^7) + 1/5^2
1/n1^2 = 1.798 x 10^6 + 0.04
1/n1^2 = 1.79804
n1^2 = 1/1.79804
n1 = √0.55608
n1 ≈ 0.746
Since the principal quantum number must be a positive integer, we round up to the nearest whole number, making the principal quantum number of the lower energy state approximately 1.