$(a)\;-\large\frac{3}{2}E_2\qquad(b)\;-\large\frac{2}{3}E_2\qquad(c)\;-\large\frac{4}{9}E_2\qquad(d)\;-\large\frac{16}{9}E_2$

Energy of electron in $n^{th}$ state

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

$E_2(H) = -\large\frac{13.6}{1}eV$

$E_3(He^+) = -\large\frac{13.6\times4}{9}eV$

$\large\frac{E_2}{E_3} = \large\frac{9}{4}$

$E_3 = \large\frac{4}{9}E_2$

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

Hence anwer is (c)

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