Ncert Exercises & Solutions

Let \(E_{n} = \frac{- 1m e^{4}}{8 \varepsilon_{0}^{2}n^{2} h^{2}} \) be the energy of the \(n^\text{th}\) level of H-atom. If all the H-atoms are in the ground state and radiation of frequency \(\frac{\left(\right. E_{2} - E_{1} \left.\right)}{h}\) falls on it, then:

(a)	it will not be absorbed at all.
(b)	some of the atoms will move to the first excited state.
(c)	all atoms will be excited to the \(n = 2\) state.
(d)	no atoms will make a transition to the \(n = 3\) state.

Choose the correct option:

1.	(b, d)	2.	(a, d)
3.	(b, c, d)	4.	(c, d)

Hint: The excitation of the atoms depends on the energy given to them.

Explanation: When all the H-atoms are in the ground state and radiation of frequency \(\frac{\left(E_2-E_1\right)}{h}\) falls on it, some of the atoms will move to the first excited state and no atoms will make a transition to the \(n = 3\) state.
Let \(E_2\) and \(E_1\) be the energy corresponding to \(n=2\) and \(n=1\) respectively. If radiation of energy \(\Delta E=\left(E_2-\right.\left.{E}_1\right)= hf\) incident on a sample where all the H -atoms are in the ground state, according to the Bohr model some of the atoms will move to the first excited state. As this energy is not sufficient for the transition from \({n}=1~\text{to}~{n}=3,\) hence no atoms will make a transition to the \({n}=3\) state.
Hence, option (1) is the correct answer.

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