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Let $\mathrm{E}_{\mathrm{n}}=\frac{-l m e^4}{8 \varepsilon_0^2 n^2 h^2}$ be the energy of the $\mathrm{nth}$ level of $\mathrm{H}-$ atom. If all the $\mathrm{H}$ - atoms are in the ground state and radiation of frequency $\frac{\left(E_2-E_1\right)}{h}$ falls on it,
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The correct answers are:
some of atoms will move to the first excited state
,
no atoms will make a transition to the $n=3$ state
some of atoms will move to the first excited state
,
no atoms will make a transition to the $n=3$ state
When all the $\mathrm{H}-$ atoms are in the ground state and radiation of photons is $\left(\mathrm{E}_2-\mathrm{E}_1\right)$ an electron jumpts to next orbit then electron jumps in next energy level $(n=2)$ after receiving this energy equal to the $\left(E_2-E_1\right)$ energy. The new state is its unstable state - electron jumps from $\mathrm{E}_2$ to $\mathrm{E}_1$ by radiating the energy of same frequency to $\left(\mathrm{E}_2-\mathrm{E}_1\right)$ So, some of atoms will move to the first excited state and no atoms will make a transition to the $\mathrm{n}=3$ state.
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