Question

Calculate the wavenumber and wavelength of the third in the brackett series of hydrogen spectrum

Answer 1

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)

λ

1

=R

H

(

n

i

2

1

−

n

f

2

1

)

Where,

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

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

H

=Rydberg’s Constant=1.09737×10

7

m

−1

n_fn

f

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

n_in

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)

λ

1

=1.09737×10

7

m

−1

(

4

2

1

−

7

2

1

)

\lambda=2164.9nmλ=2164.9nm