Question

2. The limit of Balmer series is 3646 A.
The wavelength
of first series of this
element will be​

Answer 1

Answer:

For Balmer series, n

1

=2 and n

2

=3,4,.........,∞

Hence, the wavelengths of Balmer series are given by the formula

λ

1

=R[

2

2

1

−

n

2

2

1

]

For series limit, n

2

=∞ and λ=3646A

∘

Therefore,

3646

1

=R[

2

2

1

−

∞

2

1

]

3646

1

=R[

4

1

]

R=

3646

4

Now, for first line, n

2

=3

λ

1

=R[

2

2

1

−

3

2

1

]

λ

1

=R[

4

1

−

9

1

]

λ

1

=R[

36

5

]

λ

1

=[

3646

4

][

36

5

]

λ

1

=

131256

20

λ=

20

131256

λ=6562.8A

∘