Question

Number of He ( ^{+} ) ions in 1 mole ( =6.022 times 10^{23} ) Number of ( mathrm{He}^{+} ) ions in ( 3^{mathrm{rd}} ) energy level ( =(50 / 100) times 6.022 times 10^{23} ) ( =3.011 times 10^{23} )
Number of ( mathrm{He}^{+} ) ions in ( 2^{text {nd }} ) energy level ( =(25 / 100) times 6.022 times 10^{23} ) ( =1.505 times 10^{23} )
Number of ( mathrm{He}^{+} ) ions in ( mathrm{I}^{mathrm{st}} ) enrgy level ( =(25 / 100) times 6.022 times 10^{23} )
( =1.505 times 10^{23} )
Now, ( mathrm{E}_{1} ) for ( mathrm{He}^{+} ) ion ( =mathrm{E}_{1} ) for ( mathrm{H} ) atom ( mathrm{x} mathrm{Z}^{2}(mathrm{Z} text { is the atomic number of } mathrm{He}) ) ( =-13.6 mathrm{eV} times 2^{2} )
( =-54.4 mathrm{eV} )
According to the question, Total energy released = energy released for transition from ( n=3 ) to ( n=1+ ) energy released for transition from ( n=2 ) to ( n=1 ) ( =left(E_{3}-E_{1}right) times 3.011 times 10^{23}+left(E_{2}-E_{1}right) times 1.505 times 10^{23} )
Now, ( E_{3}=E_{1} / n^{2} )
( =-54.4 / 3^{2} )
( =-6.044 mathrm{eV} )
( E_{2}=E_{1} / n^{2} )
( =-54.4 / 2^{2} )
( =-13.6 mathrm{eV} )
So, total energy released ( =(-6.044-(-54.4)) times 3.011 times 10^{23}+(-13.6-(-54.4)) times 1.505 times 10^{23} )
( =1.46 times 10^{25}+0.61 times 10^{25} )
( =2.07 times 10^{25} mathrm{eV} times 1.602 times 10^{-19} mathrm{J} )
( =3.316 times 10^{6} mathrm{J} )
( =3.316 times 10^{3} mathrm{kJ} )

# One mole of He ions is excited. An analysis showed that 50% of ions are in the third energy level. 25% are in the second energy level, and the remaining are in the first energy level. Calculate the energy emitted in kilojoules when all the ions return to the ground state.

Solution