Hello Aspirant,

Agreeing Rydberg's Theory,

1/λ = RZ² (1/n1 ² - 1/n²) ,

In the event of Shortest wavelength in Balmer's arrangement , just when change happens limitlessness to n = 2 .I mean, n1 = 2 and n = ∞

Here , λ = x , Z = 2 [ for He+, Z = 2 ]

∴ 1/x = R(2)² [ 1/2² - 1/∞² ] = R

x = 1/R - (1)

For finding longest wavelength in Paschen's arrangement.

Take n1 = 3 and n = 4

If there should be an occurrence of Li²+ , Z = 3

∴ 1/λ = R(3)² [1/3² - 1/4²]

1/λ = 9R [1/9 - 1/16] = R × 7/16

λ = 16/7R - (2)

Separating (2) ÷ (1)

λ/x = (16/7R)/(1/R) = 16/7

λ = 16x/7

Consequently, longest wavelength in the Paschen's arrangement of Li²+ is 16x/7