Question

54. The shortest wavelength of He+ ion spectrum in Balmer series is x, then longest wavelength in the Paschen series of Li+2 spectrum is :-
(1)
36x 5
(2)
16x 7
(3)
9x 5
(4)
5x 9

Answer 1

According Rydberg's Theory,
1/λ = RZ² (1/n₁² - 1/n₂²) ,
In case of Shortest wavelength in Balmer's series , only when transition takes place infinity to n = 2 .I mean, n₁ = 2 and n₂ = ∞
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 series.
Take n₁ = 3 and n₂ = 4
In case 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)

Dividing (2) ÷ (1)
λ/x = (16/7R)/(1/R) = 16/7
λ = 16x/7

Hence, longest wavelength in the Paschen's series of Li²⁺ is 16x/7
So, option (2) is correct.

Answer 2

Answer:

option B is correct answer

Explanation:

explanation is already given in above answer