Question

Potential energy is -27.7ev in second orbit of he+ then calculate double of that energy in first excited state of hydrogen atom

Answer 1

(Ans is given as -6.8eV)

Answer 2

Potential energy is -27.2 eV in second orbit of He+ then calculate double of total energy in first excited state of hydrogen atom.

A- 13.6 eV

B- 54.4 eV

C- 6.8 eV

D- 27.2 eV

Solution

Energy(Helium) = −13.6×4n2

PE=−2KE.

TE=PE2=−27.22×2=−6.8 eV

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