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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. |
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Answer» 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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