Question

Calculate the shortest wavelength in the Paschen series if the longest wavelength in the Balmar series is 6563 A^o. 

Answer 1

Given:

(λ_B) = 6563 Å = 6563 × 10⁻¹⁰ m

= 6.563 × 10⁻⁷ m 

To find: Shortest wavelength (λ_p) 

Formula: `1/lambda = "R"[1/"n"^2 - 1/"m"^2]`

Calculation: 

For (λ_B), 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 (λ_p),

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`

∴ λ_p = 8203.75 Å

The shortest wavelength in the Paschen series is 8203.75 Å.