Question
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [1/32 – 1/n2]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Solution 1
Wavelength of transition = 1285 nm
= 1285 × 10–9 m (Given)
`v = 3.29 xx 10^(15)(1/3^2 - 1/n^2)` (Given)
Since `v = c/lambda`
`= (3.0 xx 10^8)/(1285xx10^(-9)m)`
ν = 2.33 × 1014 s–1
Substituting the value of ν in the given expression,
`3.29 xx 10^(15)(1/9 - 1/n^2) = 2.33 xx 10^14`
`1/9 - 1/n^2 = (2.33xx10^14)/(3.29xx10^15)`
`1/9 - 0.7082 xx 10^(-1) = 1/n^2 `
`=> 1/n^2 = 1.1 xx 10^(-1) - 0.7082 xx 10^(-1)`
`1/n^2 = 4.029 xx 10^(-2)`
`n = sqrt(1/(4.029 xx 10^(-2)))`
n = 4.98
n ≈ 5
Hence, for the transition to be observed at 1285 nm, n = 5.
The spectrum lies in the infra-red region.
Solution 2
ν = c/λ = 3.0×108 ms-1/1285×10-9 m = 3.29×1015 (1/32 - 1/n2)
⇒ 1/n2 = 1/9 - (3.0×108 ms-1/1285×10-9 m)×(1/3.29×1015) = 0.111-0.071 = 0.04 = 1/25
⇒ n2 = 25
⇒ n = 5
The radiation corresponding to 1285 nm lies in the infrared region.
