Question

The first member of the Lyman series of H-atom has a wavelength of 1216 A. Calculate the wavelength of the second member.​

Answer 1

For Lyman series n

1

=1 and n

2

=2,3,...........,∞

Hence, the wavelength for hydrogen atom is given by the formula

λ

H

1

=RZ

H

2

[

1

2

1

−

n

2

2

1

]

1216

1

=R[

1

2

1

−

n

2

2

1

] ...................(since, Z = 1)

λ

Na

1

=RZ

Na

2

[1−

n

2

2

1

]

λ

Na

1

=R(11)

2

[

1

2

1

−

n

2

2

1

]

Therefore,

1216

1

λ

Na

1

=

R[

1

2

1

−

n

2

2

1

]

R(11)

2

[

1

2

1

−

n

2

2

1

]

λ

Na

=

(11)

2

1216

λ

Na

=

121

1216

λ

Na

=10.04≈10A

∘