Question

in a hydrogen atom an electron jumps from fourth orbit to first orbit . find out the frequency of the spectral line using Rydberg's formula ​

Answer 1

Answer:

The formula for finding the no of spectral line is N=

2

(n

2

−n

1

)(n

2

−n

1

+1)

so putting the value n

2

=4 and n

1

=1 Then N=6.

These spectral line is belong to balmer's serie.

Answer 2

Explanation:

Using the relation,

λ

1

=R(

n

1

2

1

−

n

2

2

1

)

=10

−5

(

2

2

1

−

4

2

1

)=10

−5

×(

4

1

−

16

1

)

λ=

3

16

×10

−5

cm

∴ Frequency, η=

λ

c

=

3

16

×10

−5

3×10

10

=

16

9

×10

15

Hz