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Question
Calculate the shortest wavelength in the Paschen series if the longest wavelength in the Balmar series is 6563 Ao.
Solution
Given:
(λB) = 6563 Å = 6563 × 10−10 m
= 6.563 × 10−7 m
To find: Shortest wavelength (λp)
Formula: `1/lambda = "R"[1/"n"^2 - 1/"m"^2]`
Calculation:
For (λB), m = 3, n = 2
From formula,
`1/lambda_"B" = "R"[1/2^2 - 1/3^2]`
`1/lambda_"B" = (5"R")/36`
∴ `lambda_"B" = 36/(5"R")` ....(1)
For Paschen series shortest wavelength (λp),
n = 3, m = ∞
∴ `1/lambda_"p" = "R"[1/3^2 - 1/∞]`
∴ `1/lambda_"p" = "R"[1/9]`
∴ `1/lambda_"p" = "R"/9`
∴ `lambda_"p" = 9/"R"` ....(2)
From equations (1) and (2),
`lambda_"p"/lambda_"B" = (9"/""R")/(36"/"5"R")`
∴ `lambda_"p"/lambda_"B" = 9/"R" xx (5"R")/36`
= `5/4`
∴ `lambda_"p" = 5/4 xx lambda_"B"`
= `5/4 xx 6563`
∴ λp = 8203.75 Å
The shortest wavelength in the Paschen series is 8203.75 Å.
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