Question

calculate the wavelength of the first line and the series limit for lyman series for hydrogen

Answer 1

For Lymen Series,

First line,

n₂ = 2 , n₁ = 1

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

⇒ 1/λ = R(1/1 - 1/2)

λ = 1/R(2)

⇒ λ = 911 × 2 × 10⁻¹⁰ m.

⇒ λ = 1822× 10⁻¹⁰ m.

Last line,

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

⇒ 1/λ = R(1/1 - 1/Infinity)

⇒ λ = 1/R

∴ λ = 911 Angstrom.

∴ Last line of Lyman series is 911 A°.

Hope it helps.

Answer 2

Explanation:

1/λ = R(1/1 - 1/Infinity)

⇒ λ = 1/R

∴ λ = 911 Angstrom.