Question

in h-atom electron transists from 6th orbit to 2nd orbit in multi step.then total spectral lines(without balmer series )will be

Answer 1

Explanation:

$\huge\underbrace\mathfrak\pink{Answer♡}$

n

2

=6;n

1

=2

n

2

−n

1

=4

Total spectral lines possible =

2

(n

2

−n

1

)[(n

2

−n

1

)+1]

=

2

4×5

=10

No. of Balmer lines possible ⇒

6→2;5→2;4→2;3→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