Question

S, If the total energy of an electron in the 1st shell of H atom = 0.0 eV then its potential energy in the 1st excited state would be

(a) +6.8 eV
(b) +20.4 eV
(c)-6.8 eV
(d) +3.4 eV
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Answer 1

Answer:

In 1st excited state,n=2. E=-13.6×Z^2÷n^2. E=-13.6×1^2÷2^2 (Z of H=1). E=-3.4ev. Ep=-2Ek. Ep=-2(-3.4)=+6.8ev

Answer 2

Answer:

Z=1

n=2

E= - 13.6 × Z² / n²

= - 13.6 × (1)² / (2)²

= -3.4

Ep= -2 Ek

= -2(-3.4)

= +6.8 eV