Q. 143.8( 58 Votes )

How much energy i

Answer :

The expression of energy is given by,

En = - [2.18× 10-18] Z2/n2


Where,


Z = atomic number of the atom


N = principal quantum number


For ionization from n1 = 5 to n2 = ∞,


Therefore ∆E = E2-E1 = -21.8× 10-19× [1/n22-1/n12]


= 21.8× 10-19 × [1/n22-1/n12]


=21.8× 10-19× [1/52-1/∞]


=8.72× 10-20J


For ionization from 1st orbit, n1=1, n2=∞


Therefore ∆E’ = 21.8× 10-19× [1/12- 1/∞]


= 21.8× 10-19J


Now ∆E’/∆E = 21.8× 10-19/8.72× 10-20 = 25


Thus, the energy required to remove electron from 1st orbit is 25 times than the required to electron from 5th orbit.


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