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1g hydrogen atoms are excited by radiations. The study of spectrum indicates that 27% of the atoms are in 3rd energy level and 15% of atoms are in 2nd energy level and the rest in the ground state. Ionization potential of hydrogen is 13.6 eV. Calculate (a) number of atoms present in energy level 1st, 2nd and 3rd (b) total energy released in joules when all atoms return to ground state.

Answer» <html><body><p></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> :By mole concept, 1g of hydrogen atoms <a href="https://interviewquestions.tuteehub.com/tag/contain-409810" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAIN">CONTAIN</a> `= 6.02 xx 10^(23)` atoms <br/> `:.` 1.8g of hydrogen atoms contain <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>-atom `= 6.02 xx 10^(23) xx 1.8 = 10.84 xx 10^(23)` <br/> Atoms present in 3rd energy level `= (27)/(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>) xx 10.84 xx 10^(23) = 292.68 xx 10^(21)` <br/> Atoms present in 2nd energy level `= (<a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a>)/(100) xx 10.84 xx 10^(23) = 162.6 xx 10^(21)` <br/> Atoms present in 1st energy level `= (58)/(100) xx 10.84 xx 10^(23) = 628.72 xx 10^(21)` <br/> Ionization potential of H = 13.6 eV means that <br/> `E_(1) = -13.6 eV, E_(2) = - (13.6)/(2^(2)) eV, E_(3) = - (13.3)/(3^(2)) eV` <br/> Energy released from all atoms when electrons return from 3rd level to 1st level <br/> `=(E_(3) - E_(1)) xx 292.68 xx 10^(21) = (-(13.6)/(9) + 13.6) xx 292.68 xx 10^(21) eV` <br/> `= 3.537 xx 10^(24) eV` <br/> Energy released from all atoms when electrons return from 2nd level to 1st level <br/> `= (E_(2) - E_(1)) xx 162.6 xx 10^(21) = ((-13.6)/(4) + 13.6) xx 162.6 xx 10^(21) eV` <br/> `= 1.659 xx 10^(24) eV` <br/> `:.` Total energy released `= (3.537 + 1.659) xx 10^(24) eV` <br/> `= 5.196 xx 10^(24) eV = (5.196 xx 10^(24)) xx (1.602 xx 10^(19)) J` <br/> `= 8.3239 xx 10^(5) J = 832.4 kJ`</body></html>


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