Question

What is the shortest wavelength line in the paschen series of Li2+ ion
(1.).1/R
(2)...3/R
(3)....1/3R
(4).....4R

Answer 1

Answer ⇒ No option is correct.

The shortest wavelength of the light emitted is when the electron jumps from Infinity to 3.

For Paschen Series,

n₂ = Infinity , n₁ = 3

∴ $\frac{1}{\lambda} = R_H (\frac{1}{(n_1)^2} - \frac{1}{(n_2)^2} )$

⇒ 1/λ = R(1/9 - 1/Infinity)

λ = 1/R × (9)

or λ = 9/R

For Li²⁺, λ = 9/R × 3² m. = 81/R m.

This is the Required Answer.

Hope it helps.

Answer 2

Here's your answer ....

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