Question

In H-atom, electron transits from $6^{th}$ orbit to $2^{nd}$ orbit in multi step, then total spectral lines (without Balmer series) will be:

• A. $6$
• B. $10$
• C. $4$
• D. $0$

Answer 1

The correct option is A $6$

$n_2=6;n_1=2$

$n_2-n_1=4$

Total spectral lines possible $=\frac{(n_2-n_1)\left[\right(n_2-n_1)+1]}{2}$

$=\frac{4\times 5}{2}=10$

No. of Balmer lines possible $\Rightarrow$

$6\to 2;5\to 2;4\to 2;3\to 2;$

Total 4 Balmer lines arfe possible for the given transaction

Expacted spectral line= Total spectral lines possible - No. of Balmer lines possible

= $10-4$ = $6$
Hence, the correct option is 6.