A gas of identical H-like atom has some atoms in the lowest (ground) energy level A and few atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The a

Question

A gas of identical H-like atom has some atoms in the lowest (ground) energy level A and few atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level by absorbing monochromatic light of photon energy 2.55 eV . Subsequently, the atoms emit radiation of only six different photons energies. Some of the emitted photons have energy 2.55 eV . Some have more and some have less than 2.55 eV . How many statements is/are correct?

Options:

(i) The principal quantum number of initially excited level $B = 2$ .

(ii) The ionisation energy for the gas atoms is 13.6 eV .

(iii) The maximum and the minimum energies of the emitted photons will be 12.75 eV and 0.66 eV respectively.

(iv) If all emitted photons have energy $\geq 2.55 eV$ , then in this case principal quantum number of initially excited level $B = 3$ .

TestHub · Accepted Answer

Since we obtain 6 emission lines, it means electron comes from 4th orbit energy emitted is equal to, less than and more than 2.55 eV . So it can be like this :

$E_{4} - E_{2} = 2.55 eV, E_{4} - E_{3} < 2.55 eV E_{4} - E_{1} > 2.55 eV$

(i) $n = 2$ ,

$(E_{4} - E_{2})^{atom} = (E_{4} - E_{2})^{H} \times Z^{2} 2.55 = 2.55 \times Z^{2}$

$Z = 1$

(ii) IP=13.6 $Z^{2} = 13.6 \times (1)^{2} = 13.6 eV$

(iii) Maximum energy emitted $= E_{4} - E_{1}$

$= (E_{4} - E_{1})^{H} \times Z^{2}$

$= 12.75 \times (1)^{2}$

$= 12.75 eV$

$M inim u m e n er g ye mi tt e d = E_{4} - E_{3}$

$= (E_{4} - E_{3})^{H} \times Z^{2} = 0.66 \times (1)^{2} = 0.66 eV$

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