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Question
Calculate the longest wavelength in the Paschen series.
(Given RH =1.097 ×107 m-1)
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Solution
Given: n = 3, m = 4
To find: Longest wavelength in Paschen series (λL)
Formula: `1/lambda_"L" = "R"[1/"n"^2 - 1/"m"^2]`
Calculation:
From formula,
`1/lambda_"L" = "R"[1/3^2 - 1/4^2]`
∴ `1/lambda_"L" = "R"[1/9 - 1/16]`
= `"R"[(16 - 9)/(9 xx 16)] = (1.097 xx 10^7 xx 7)/(9 xx 16)`
∴ `lambda_"L" = (9 xx 16)/(1.097 xx 7) xx 10^-7`
= antilog {log(9) + log(16) − log(1.097) − log(7)} × 10−7
= antilog{0.9542 + 1.2041 − 0.0402 − 0.8451} × 10−7
= antilog{1.2730} × 10−7
= 18.75 × 10−7 m
∴ λL = 18750 Å
The longest wavelength in the Paschen series is 18750 Å.
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