Let `E_(n)=(-1-me^(4))/(8epsi_(0)^(2)n^(2)h^(2))` be the enrgy 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

Options

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. no atoms will make a transition to the n = 3 state.

Answer

B, D

Explanation

When photons with frequency `(E_(2)-E_(1))/(h)` incident on the given hydrogen atoms lying in ground state n = 1, because of continuous orbital motion of electrons in these atoms, all the incident photons are not able to make collisions with the electrons. Here whichever electrons happen to collide with incident photons, have their final energy equal to
`E_(2).=E_(1)+hv :.E_(2).=E_(1)+h((E_(2)-E_(1))/(h))=E_(2)`
Hence all such electrons go into n = 2 state i.e. first excited state.
`rArr` Thus, option (B) is correct. Here energy obtained from incident radiation `=E_(2)-E_(1)=-3.4-(-12.6)=10.2eV` which is not sufficient to take the electron into n = 3 state. Thus, none of given hydrogen atoms would go into n = 3 state.
`rArr` Thus, option (D) is correct.