Question

calculate the wavelength of first spectral of Balmer series for hydrogen. Given that Rydberg constant ,R = 1.097 × 10^7 m^-1.​

Answer 1

The wavelength of photon emitted due to the transition from n to m state of a hydrogen atom is given by

λ

1

=R

H

(

n

2

1

−

m

2

1

)

Let the wavelengths of first and second line of Balmer series be λ

1

and λ

2

respectively.

Now for first wavelength of Balmer series, n=2 and m=3

∴

λ

1

1

=1.097×10

7

(

2

2

1

−

3

2

1

)⟹λ

1

≈6563A

o

For second wavelength of Balmer series, n=2 and m=4

∴

λ

2

1

=1.097×10

7

(

2

2

1

−

4

2

1

)⟹λ

2

≈4861A

o