Q1 / 160 · AP EAMCET 2024 (21 May Shift 1) · Chemistry

The energy of third orbit of $\mathrm{Li}^{2+}$ ion (in J ) is

  • Correct answer
    A
    $-2.18 \times 10^{-18}$
  • B
    $-6.54 \times 10^{-18}$
  • C
    $-7.3 \times 10^{-19}$
  • D
    $+2.18 \times 10^{-18}$

Solution

Energy $\left(\mathrm{E}_{\mathrm{n}}\right)=-2.18 \times 10^{-18}\left(\frac{\mathrm{z}^2}{\mathrm{n}^2}\right) \mathrm{J}$ $Z=3$ $\mathrm{n}=3$ (third orbit) $E_3=-2.18 \times 10^{-18} \times \frac{(3)^2}{(3)^2} J$ $E_3=-2.18 \times 10^{-18} \mathrm{~J}$

Hence, the correct answer is option: 'A'.

Relevant Questions