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A gas of identical hydrogen-like atoms h...

A gas of identical hydrogen-like atoms has some atoms in the lowest in lower (ground) energy level `A` and some atoms in a partical upper (excited) energy level `B` and there are no atoms in any other energy level.The atoms of the gas make transition to higher energy level by absorbing monochromatic light of photon energy `2.7 e V`.
Subsequenty , the atom emit radiation of only six different photon energies. Some of the emitted photons have energy `2.7 e V` some have energy more , and some have less than `2.7 e V`.
a Find the principal quantum number of the intially excited level `B`
b Find the ionization energy for the gas atoms.
c Find the maximum and the minimum energies of the emitted photons.

Text Solution

Verified by Experts

a. Since only six different transition take place , the final state is `n = 4`. The energy level of hydrogen atom are given by
`E_(n) = - (13.6)/(4^(2)) e V`
If `n_(B)` is the principal quantum number of the initially excited state `B` , than
`E_(4) - E_(n_B) = - (13.6)/(4^(2)) - ((- 13.6)/(n_(B)^(2)))`
` = 13.6 [(1)/(n_(B)^(2)) - (1)/(16)]`
`E_(4) - E_(n_B) = 2.7 e V`
`2.7 = 13.6 [(1)/(n_(B)^(2)) - (1)/(16)]`
which gives `n_(B) = 2`. ("Rounding off to nearst integet")`
b. The transition energy is numerically equal to the ground state energy `E_(1)` of level A.
`E_(4) = (E_(1))/(16), E_(2) = (E_(1))/(4)`
`E_(4) - (E_(2) = E_(1))/(16) - (E_(1))/(4)`
`2.7 e V = - (3)/(16) E_(1)`
`E_(1) = - 14.4 e V`
Thus , the ionization energy of the given atom is `14.4 e V`
c. Maximum energy of the emitted photon is for the electron transition `n = 4` to `n = 1`, i.e.,
`E_(4) - E_(3) = (E_(1))/(16) - E_(1) = - (15)/(16) E_(1)`
`= (15)/(16) xx (14.4) = 13.5 e V`
Thus, the maximum energy of the emitted photon is `13.5 e V`.
Maximum energy of the emitted photon corresponds to the transition `n = 4 to n = 3`, i.e.,
`E_(4) - E_(3) = (E_(1))/(16) - E_(1)/(9) = - (7)/(144) E_(1)`
`= - (7)/(144) xx (-14.4) = 0.7 e V`
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