Atoms And Nuclei Question 265

Question: The energy of $ H{{e}^{+}} $ in the ground state is -54.4 eV, then the energy of $ L{{i}^{++}} $ 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

    $ E_{n}=-( Rch )\frac{Z^{2}}{n^{2}}=-54.4eV $

    For $ H{{e}^{+}} $ is ground state $ E_{1}=-(Rch)\frac{{{(2)}^{2}}}{{{(1)}^{2}}}=-54.4\Rightarrow Rch=13.6 $
    $ \therefore $ For $ L{{i}^{++}} $ in first excited state (n=2) $ E=-13.6\times \frac{{{( 3 )}^{2}}}{{{( 2 )}^{2}}}=-30.6eV $



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