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A singly ionized helium atom in an excit...

A singly ionized helium atom in an excited state (n = 4) emits a photon of energy 2.6 eV. Given that the ground state energy of hydrogen atom is 13.6 eV, the energy `(E_t)` and quantum number (n) of the resulting state are respectively,

A

`E_(t)= -13.6 eV, n=1`

B

`E_(t)= -6.0 eV, n= 3`

C

`E_(t)= 6.0 eV, n= 2`

D

`E_(t)= -13.6 eV, n=2`

Text Solution

Verified by Experts

The correct Answer is:
B
`2.6 = 13.6Z^(2) [(1)/(n^(2))- (1)/(4^(2))] rArr (2.6)/(13.6 xx 4)= [(1)/(n^(2))- (1)/(4^(2))]`
`rArr (2.6)/(13.6 xx 4)= (1)/(n^(2))- (1)/(16) rArr n= 3`
Now, energy `E= (13.6Z^(2))/(n^(2)) eV rArr - (13.6 xx 4)/(9) eV= - 6eV`
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