Atoms And Nuclei Question 265

Question: The energy of He+ in the ground state is -54.4 eV, then the energy of Li++ in the first excited state will be

Options:

A) -30.6 eV

B) 27.2 eV

C) -13.6 eV

D) - 27.2 eV

Show Answer

Answer:

Correct Answer: A

Solution:

  • Energy of electron in nth orbit is

    En=(Rch)Z2n2=54.4eV

    For He+ is ground state E1=(Rch)(2)2(1)2=54.4Rch=13.6
    For Li++ in first excited state (n=2) E=13.6×(3)2(2)2=30.6eV



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