MCQ
The total energy of the electron of $H-$ atom in the second quantum state is $-E_2$. The total energy of the $He^+$ atom in the third quantum state is
  • A
    $ - \frac{3}{2}{E_2}$
  • B
    $ - \frac{2}{3}{E_2}$
  • C
    $ - \frac{16}{9}{E_2}$
  • $ - \frac{4}{9}{E_2}$

Answer

Correct option: D.
$ - \frac{4}{9}{E_2}$
d
Energy of electron in nth state

$=\frac{- Z ^2 \times 13.6\, eV }{ n ^2}$

$E _2( H )=-13.6 / 1 \,eV$

$E _3\left( He ^{+}\right)=-13.6 \times \frac{4}{9}\, eV$

$E _2 / E _3=-9 / 4$

$E _3=-\frac{4}{9} \,E _2$

For negative value of $E_2, E_3$ will also be negative.

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