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The ionisation energy of hydrogen is 13....

The ionisation energy of hydrogen is 13.6 eV . The energy of the photon released when an electron jumps from the first excited state (n=2) to the ground state of hydrogen atom is

A

3.4 eV

B

4.53 eV

C

10.2 eV

D

13.6 eV

Text Solution

Verified by Experts

The correct Answer is:
C
Energy of the electron in `n^(th)` state of hydrogen atom is
`" "E_(n)=-13.6/n^(2) eV`
For ground state, n = 1
`therefore E_(1)=-(13.6)/1^(2)eV`
For first excited state, n = 2
`therefore E_(2)=-13.6/2^(2)eV`
The energy of the photon emitted for the electron transition from n = 2 to n = 1 is
`DeltaE = E_(2)-E_(1)=(-13.6 eV)(1/2^(2)-1/1^(2))=10.2eV`
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