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The ionization energy of the ground stat...

The ionization energy of the ground state of hydrogen atom is `2.18 xx 10^(-8)J`. The energy of an electron in its second orbit would be

A

`-1.09 xx 10^(-18)` J

B

`-2.18 xx 10^(-18)` J

C

`-4.36 xx 10^(-18)` J

D

`-5.45 xx 10^(-19)` J

Text Solution

Verified by Experts

The correct Answer is:
D
`E_(n) = (-E_1)/(n^2)`
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