Energy of electron in first excited state in Hydrogen atom is -3.4eV.

1.	Energy of electron in first excited state in Hydrogen atom is -3.4eV. Find KE and PE of electron in the ground state.
Answer» Energy of electron in n = 2 is -3.4eV ∴ energy in ground state = -13.6eV kE = -TE = +13.6eV E_n = x/n² = - 3.4 eV = x/ 2² energy in ground state x = - 13.6eV. PE = 2TE = -2×13.6eV = -27.2eV

Energy of electron in first excited state in Hydrogen atom is -3.4eV. Find KE and PE of electron in the ground state.

Answer»

Energy of electron in n = 2 is -3.4eV

∴ energy in ground state = -13.6eV

kE = -TE = +13.6eV

E_n = x/n²

= - 3.4 eV = x/ 2²

energy in ground state x = - 13.6eV.

PE = 2TE = -2×13.6eV = -27.2eV

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Energy of electron in first excited state in Hydrogen atom is -3.4eV. Find KE and PE of electron in the ground state.

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