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The ratio of energy of the electron in g...

The ratio of energy of the electron in ground state of hydrogen to the electron in first excited state of `Be^(3+)` is

A

`1:4`

B

`1:8`

C

`1:16`

D

`16:1`

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(n) = (-13.6 Z^(2))/(n^(2)) eV "atom"^(-1)`
For hydrogen in ground state
`E_(1) = -13.6 eV "atom"^(-1)` (Z = 1, n =1)
For `Be^(3+)` in first excited state
`E_(2) = (-13.6 xx 4^(2))/(2^(2)) eV "atom"^(-1)`
`= -13.6 xx 4 eV "atom"^(-1)`
`E_(1) : E_(2) = 1:4`
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