Question

If there are only two $H-$atoms, each is in $3^{rd}$ excited state then:

• A. maximum number of different photons emitted is $4$
• B. maximum number of different photons emitted is $3$
• C. minimum number of different photons emitted is $1$
• D. minimum number of different photons emitted is $2$

Answer 1

Answer: (A)

For $n^{th}$ exited state $n=(n^{th}+1)$

Here $n^{th}=3^{rd}=(3+1)$

$n=4$

For the first $H-$atom the possible no.of photons emitted $\Rightarrow 4\to 3$

$3\to 2$

$2\to 1$

And

For the second $H-$atoms, the possible no. of photons emitted $\Rightarrow 4\to 2$

$[2\to 1]$ (repeated)

Therefore, the maximum no. of photons emitted $\Rightarrow 4$