In a sample containing 3 hydrogen atoms excited to the 4th
orbit, the maximum and the minimum number of lines that will
be obtained on de-excitation are respectively
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Answer:
2:1
Balmer series→Visible→n=2
lyman series→Ultraviolet→n=1
n
2
=4,given
No of spectral lines in visible region=
2
(n
2
−n
1
)(n
2
−n
1
+1)
=
2
(4−2)(4−2+1)
=3
No of spectral lines in ultraviolet region=
2
(n
2
−n
1
)(n
2
−n
1
+1)
=
2
(4−1)(4−1+1)
=6
ratio of spectral lines in UV and visible 6:3=2:1
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