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if potential energy of an electron in a hydrogen atom in first excited state is taken to be zero ,Kinetic energy(in eV) of an electron in ground state will be?

if potential energy of an electron in a hydrogen atom in first excited state is taken to be zero ,Kinetic energy(in eV) of an electron in ground state will be?

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4 Answers

Abhishek Kumar
askIITians Faculty 253 Points
6 years ago
TE in first excited state = -13.6/4 = -3.4 eV
PE in the first excited state = -6.8 eV (-PE = 2 X -TE)
For this to be zero, We must add 6.8 eV. So, add this 6.8 eV to all energy values to get the new values because shifting of zero or reference simply shifts all energy by the same value.
KE in ground state = 13.6 eV
So, in new system = 13.6 + 6.8 = 20.4 eV
Arshdeep Singh
11 Points
4 years ago
The Kinetic Energy is independent of the position of frame of reference i.e. where the potention energy will be zero. So whether we take p.e. zero at infinity or at ground state the value of kinetic energy of a election moving in a particular orbit will remain constant to move in the same orbit. Thus the kinetic energy will remain equal to 13.6 eV in the ground state.
Ritu Bisht
66 Points
4 years ago
If zero of potential energy is changed, KE does not change and continues to be + 13.6 eV.KE is independent of frame of reference of potential energy
 
ankit singh
askIITians Faculty 614 Points
one year ago

For an electron:
U=2KE
E=U+KE=2U
Energy in first excited state, n=2
E(n=1)=13.6 eV
U(n=1)=2E=27.2 eV
E(n=2)=n213.6=3.4 eV
If U(n=1) is taken as the reference value
E(n=2)=E(n=2)U(n=1)=23.8 eV

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