Question

Calculate the wavelength of third line in brackett series for hydrogen

Answer 1

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

∴ 1/λ = R(1/4² - 1/7²)

⇒ 1/λ = R(1/16 - 1/49)

⇒ 1/λ = R(0.04209)

⇒ 1/λ = R × 10.04209

λ = 911/10.04209

λ = 21643.15 Angstrom.

Hope it helps.

Answer 2

Answer: The wavelength of the line will be  2164.9 nm

Explanation: Using Rydberg's Equation:

$\frac{1}{\lambda }=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right)$

Where,

$\lambda=\text{Wavelength of radiation}=?nm$

$R_H=\text{Rydberg's Constant}=1.09737\times 10^7m^{-1}$

$n_f$ = Higher energy level = 7 (Third line from Brackett series)

$n_i$ = Lower energy level = 4 (Brackett series)

Putting the values, in above equation, we get

$\frac{1}{\lambda }=1.09737\times 10^7m^{-1}\left(\frac{1}{4^2}-\frac{1}{7^2} \right)$

$\lambda=2164.9nm$