1.6g sample of He​ions are excited to emit radiations . Spectral study of sample indicates that highest excited state is 2nd and that total no of ions in 1st and 2nd​ excited state is one third of ions in ground state.Total energy evolved on de-excitation of sample is 429.5kJ.Let X be the percentage of He ions in second excited stage in original sample .Find the value of [X/6.25]

Dear Student,

Moles of He+=MassAtomic mass                       =1.64                       =0.04 molNow, 1 mol of He+ contains 6.022×1023 ionsSo, 0.04 mol of He+ contains 0.04×6.022×1023 =2.408×1022 ionsHence, total no. of ions =2.408×1022 ionsLet a be the total ions in 2nd excited state and b be the ions in 1st excited state. c is the ions in ground state. So, 13(a+b)=cor, 13a+13b-c = 0      ...(i)also, a+b+c=2.408×1022  ..(ii)On adding these two equation, we can find the value of a+b and then label the equation (iii)When all the excited ions return to normal state, the energy released= Energy released during deexcitation from II to I+ Energy released during deexcitation from III to Ior, 429.5 × 103 J = [E2-E1]×b + [E3-E1]×aor, 429.5 × 103 = [(-13.622×22-(-13.612×22)]×b + [(-13.632×22-(-13.612×22)]×a On solving these two, one more equation in terms of a+b will be formed which is labelled (iv)and, solving (iii) and (iv) we can get values of a and b and can eventually find the value of X. 

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