Question

Let $E_n=\frac{-1}{8{\epsilon }_0^2}\frac{m{e}^4}{n^2{h}^2}$ be the energy of the $n^{th}$ level of $H$- atom. If all the $H$- atoms are in the ground state and radiation of frequency $(E_2–E_1)/h$ falls on it.

• A. no atoms will make a transition to the $n=3$ state.
• B. all atoms will be excited to the $n=2$ state.
  
• C. it will not be absorbed at all
  
• D. some of atoms will move to the first excited state.
  

Answer 1

Answer: (A), (D)

some of atoms will move to the first excited state.


Given, energy of the$n^{th}$ level of $H$-atom, $E_n=\frac{-1}{8{\epsilon }_0^2}\frac{m{e}^4}{n^2{h}^2}$

Then $E_2$ and$E_1$ be the energy corresponding to $n=2$ and $n=1$ level of $H$-atom respectively.

If radiation of energy $\Delta E=(E_2–E_1)=hf$ incident on a sample where all the $H$-atoms are in the ground state, according to Bohr’s model, some of the atoms will move to the first excited state.

As this energy is not sufficient for transition from $n=1$ to $n=3$, hence no atoms will make a transition to the $n=3$ state.

$\text{Final answer}:$ (b), (d)