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The energy of an electron in the first B...

The energy of an electron in the first Bohr orbit of H atom is `-13.6 eV`. The possible energy values (s) of the excited state (s) for electron in bohr orbits of hydrogen is (are)

A

`-3.4eV`

B

`-4.2eV`

C

`-6.8eV`

D

`+6.8eV`

Text Solution

Verified by Experts

The correct Answer is:
A
The energy of an electron in a Bohr atom is expressed as `E_(n)=(kz^(2))/(n^(2))`
where, k = Constant, `=-13.6eV" for "H(n=1)`
Z = Atomic number,
n = Orbit numberwhen `n=2, E=(-13.6)/(2^(2))eV=-3.40eV`
(n can have only integral value `1, 2, 3,…….oo` )
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