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Question: Let \(E_{n} = \frac{- me^{4}}{8\varepsilon_{0}^{2}n^{2}}\)be the energy of the \(n^{th}\) level of H...

Let En=−me48ε02n2E_{n} = \frac{- me^{4}}{8\varepsilon_{0}^{2}n^{2}}be the energy of the nthn^{th} level of H-atom. If all the H-atoms are in the ground state and radiation of frequency(E2−E1)/h(E_{2} - E_{1})\text{/h} falls on it, then

A

It will not be absorbed at all

B

Some of atoms will move to the first excited state

C

All atoms will be excited to the n = 2 state

D

All atoms will make a transition to the n = 3 State

Answer

Some of atoms will move to the first excited state

Explanation

Solution

The given energy of nth level of hydrogen atom is En=−me48ε02n2h2E_{n} = - \frac{me^{4}}{8\varepsilon_{0}^{2}n^{2}h^{2}}

Since all the H atom are in ground state (n = 1) then the radiation of given frequency E2−E1h\frac{E_{2} - E_{1}}{h} falling on it may be absorbed by some of the atoms and move them to the first excited state (n = 2)