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.
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
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