Calculate the ratio of energy of third line of balmer series and sixth line of layman series
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Explanation:
Lyman series, n
1
=1
and for the third line, n
2
=4
v
1
=R
H
Z
2
(
n
1
2
1
−
n
2
2
1
)
=R
H
(1)
2
(
1
1
−
16
1
)=
16
15
R
H
(∵ For H atom Z=1)
For Balmer series, n
1
=2 and for first line, n
2
=3
v
2
=R
H
Z
2
(
(2)
2
1
−
(3)
2
1
) [ForLi
2+
,Z=3]
=R
H
(3)
2
(
4
1
−
9
1
)=9R
H
(
36
5
)
v
2
v
1
=
16×9×5R
H
15R
H
×36
=
4
3
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