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A gas of identical H-like atom has someatoms in the lowest (ground) energy level A and some 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 aborbing monochromatic light of photon energy 2.7eV. Subsequently, the atoms emit radiation of only six different photons energies. Some of the emitted photons have energy 2.7eV. Some have more and some have less than 2.7eV. (a) Find theprincipal quantum number of initially excitied level B. (b) Find the ionisation energy for the gas atoms. (c ) Find the maximum and the minimum energies of the emitted photons.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Since we obtain 6 emission lines, it means <a href="https://interviewquestions.tuteehub.com/tag/electron-968715" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRON">ELECTRON</a> comes from <a href="https://interviewquestions.tuteehub.com/tag/4th-319068" style="font-weight:bold;" target="_blank" title="Click to know more about 4TH">4TH</a> orbit energy emitted is equal to less than and more than `2.7eV`. So it can be like this<br/> `E_(4)-E_(2)=2.7eV,""E_(4)-E_(3)lt2.7eV`, <br/> `E_(4)-E_(1)gt2.7eV` <br/> (a) `n=2` <br/> `(E_(4)-E_(2))^("atom")=(E_(4)-E_(2))^(H)xxZ^(2)`<br/> `2.7=2.55xxZ^(2)=1.029`<br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) `IP=13.6Z^(2)=13.6xx(1.029)^(2)=14.4eV` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> ) Maximum energy emitted `=E_(4)-E_(1)=(E_(4)-E_(1))^(H)xxZ^(2)`<br/> `=12.75xx(1.029)^(2)`<br/>`13.5eV` <br/> Minimum energy emitted `=E_(4)-E_(3)=(E_(4)-E_(3))^(H)xxZ^(2)`<br/> `=66xx(1.029)^(2)=0.7eV`</body></html>


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