Question

If potential energy of electron in the first excited state of hydrogen atom is taken as zero then the energy of first excited state and that of first line of lyman series are respectively

Answer 1

Explanation:

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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. For this to be zero, We must add 6.8 eV.

Answer 2

Answer:

energy of electron in its nth orbit = -13.6/nˆ2

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 = 20.4eV

Explanation: