the shortest wavelength of He atom in balmer series is x, what is the longest wavelength in the paschen series of Li 2+
Answer (1)
Student Expert
4th Oct, 2019
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
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