The wave length of first member of Balmer
series of a hydrogen atom is nearly (The value
of Rydberg constant R = 1.08 x 10'm-')
1) 4400A°
2) 5500A
3) 6600A
4) 7700A
Answers
Answer:
third option
Explanation:
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ans-
The first member of the Balmer series of hydrogen atom has wavelength of 656.3 nm, Calculate the wavelength and frequency of the second member of the same series. Given : C=3×10
8
ms
"1
.
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ANSWER
We have
λ
0
1
=R[
n
1
2
1
−
n
2
2
1
]
For first member of Balmer series
n
1
=2 and n
2
=3
λ
1
=6563A
∘
6563×10
−10
1
=R(
2
2
1
−
3
2
1
)=
36
5R
....(1)
For second member of Balmer series, n
1
=2 and n
2
=4
λ
2
1
=R(
4
1
−
16
1
)=
16
3R
....(2)
Dividing equation (1) by equation (2)
6563×10
−10
λ
2
=
36
5
×
3
16
λ
2
=
108
5×16×6563×10
−10
=4861 A
∘
Frequence v=
λ
2
C
=
4861×10
−10
3×10
8
=0.0006171×10
18
v=6.17×10
14
Hz.
Answer By
Sanatan singh