Question:
the ground state and the first excited state energies of hydrogen atom are -13.6eV and -3.4eV resp. If potential energy in ground state is taken to be zero, then
1) Potential Energy in the first excited state would be 20.4 eV
2) Total Energy in the first excited state would be 23.8 eV
3) Kinetic Energy in the first excited state would be 3.4 Ev
4) Total energy in the ground state would be 13.6 eV
MULTICORRECT
Answer:
Total energy of an electron in an orbit is given by
En = -13.6 / n2 eV where n indicates the orbit number
Kinetic energy = -E
Potential energy = -2KE
Case 1
Hence, for ground state, n =1, TE = -13.6 / 12 = -13.6 eV
KE = -E = -(-13.6) = 13.6 eV
PE = -2KE = -2 * 13.6 = -27.2 eV
Case 2
For first state, n =2, TE = -13.6 / 22 = -3.4 eV
KE = -E = -(-3.4) 3.4 eV
PE = -2KE = -2(3.4) = - 6.4 eV
Hence, choices for the question will be
- False ; PE of first state = -6.4 ev
- False ; TE of first state = -3.4 eV
- True
- True
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