Question

In certain electronic transition from quantum level n to ground state in atomic hydrogen in one or more steps no line belonging to Brackett series is observed. The wave numbers which may be observed in Balmer series is then?

• A. $\frac{8R}{9},\frac{5R}{36}$
• B. $\frac{3R}{16},\frac{8R}{9}$
• C. $\frac{5R}{36},\frac{3R}{16}$
• D. $\frac{3R}{4},\frac{3R}{16}$

Answer 1

The correct option is C $\frac{5R}{36},\frac{3R}{16}$

$Given:$ In certain electronic transition from quantum level n to ground state in atomic hydrogen in one or more steps no line belonging to Brackett series is observed.

$Solution:$

The $n$ should be equal to 4 ; otherwise if $n>4$,

then Bracket series will also be noticed.

Total no.of lines formed $=n(n-1)/2=4\times 3/2=6$

No. of lines formed will be due to transition

between these energy levels $n=4$ to $n=3$

$n=4\mathrm{ton}=2$

$n=4$ to $n=1$

$n=3\mathrm{ton}=2$

$n=3\mathrm{ton}=1$

$n=2$ to $n=1$

Balmer series line obtained when any electron comes to second orbit from outer orbit so

out of these six lines, only two lines belongs to Balmer series.

i.e., from $n=4$ to $n=2$ and

$n=3$ to $n=2$

For these lines wave number is,

$\left(i\right)\overline{v1}=\frac{1}{\lambda }=RH[\frac{1}{2^2}-\frac{1}{4^2}]=\frac{12}{64}RH=\frac{3}{16}RH$

$\left(ii\right)\overline{v2}=\frac{1}{\lambda }RH[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{5}{36}{R}_H$

$So,the$ $correct$ $option:C$