Question

calculate the wave length of second line in the balmer series​

Answer 1

Answer:

Correct option is

A

20

27

×4861

A

˚

For the first line in balmer series:

λ

1

=R(

2

2

1

−

3

2

1

)=

36

5R

For second balmer line:

4861

1

=R(

2

2

1

−

4

2

1

)=

16

3R

Divide both equations:

4861

λ

=

16

3R

×

5R

36

λ=4861×

20

27

Explanation:

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Answer 2

Answer:

the wavelength of the second line of the balmer series in the hydrogen spectrum is 4861 Armstrong calculate the wavelength of the first line For the first line of any series (For Balmer, n = 2), wavenumber (1/λ) is represented as: 1/λ = R [1/n² - 1/ (n+1)²], R is the Rydberg constant.

Explanation: