Question

The difference between the wave number of first line of Balmer series and the last line of paschen series for Li +2 ion is ?????

Answer 1

First line of the Balmer Series, n₂ = 2 and n₁ = 3

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

⇒ 1/λ = R(1/4 - 1/9)

∴ λ = 1/R × 36/5

⇒ λ = 911 × 10⁻¹⁰ × 36/5 m.

λ = 6559.2 × 10⁻¹⁰ m.

∴ For Li²⁺, λ = 6559.2 × 10⁻¹⁰ × 3² = 59032.8 × 10⁻¹⁰ m.

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(0.11)

⇒ λ = 8199 × 10⁻¹⁰ m.

For Li²⁺, λ = 8199 × 10⁻¹⁰ × 3² m. = 73791 × 10⁻¹⁰ m.

∴ Difference in Wave number = 1/59032.8 × 10⁻¹⁰ - 1/73791 × 10⁻¹⁰

= 10¹⁰ (0.00001693973 - 0.00001355178)

= 10¹⁰(0.00000338795)

= 33879.5 m⁻¹

Hope it helps.

Answer 2

Answer:

R/4 the difference between wave nbr