Maximum number of different photons emitted when only two H atoms are placed in a container and each is present in 3rd excited state:

Option 1 -

Option 2 -

Option 3 -

Option 4 -

2 Views | Posted 4 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

4 weeks ago

Correct Option - 4

Detailed Solution:

$1^{st} H$ - atom

$n = 4$

$n = 1$

$4 \to 3,3 \to 2,2 \to 1$

$4 \to 1$ or $4 \to 3,3 \to 1$

or $4 \to 2,2 \to 1$

Hence, number of different lines possible $= 3 + 1 = 4$

Minimally, both can have same transition i.e $4 \to 1$ etc.

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