Question

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​

Answer 1

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