# How much energy i

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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