1.

The ground state energy of hydrogen atom is -13.6 eV (i) What are the potential energy and K.E. of electron is 3rd excited state? (ii) If the electron jumped to the ground state form the third excited state, calculate the frequency of photon emitted.

Answer» Correct Answer - `(i) -1.7eV; 0.85eV ; (ii) 3xx10^(15)Hz`
Here, `E_(1)=-13.6eV`
for third excited state, n=4
`:. E_(4)=(-13.6)/(4^(2))=-0.85eV`
`:. K.E.=-E_(4)=0.85eV`
`P.E. =-2 (K.E.)=-2(0.85)eV=-1.70eV`
Energy emitted, `DeltaE=E_(4)-E_(1)`
`hv=-0.85-(-13.6)eV=12.75eV`
`v=(12.75xx1.6xx10^(-19))/(6.6xx10^(-34))=3xx10^(15) Hz`


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