The potential energy of the electron present in the ground state of \(\mathrm{Li}^{2+}\)ion is represented by:

The potential energy of the electron present in the ground state of \(\mathrm{Li}^{2+}\)ion is represented by:
  1. \(\frac{-3 e}{4 \pi \varepsilon_0 r}\)
  2. \(\frac{+3 \mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\)
  3. \(\frac{-3 \mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\)
  4. \(\frac{-9 e^2}{4 \pi \varepsilon_0 r}\)

Solution

The potential energy of an electron in the ground state of hydrogen-like atoms (one electron system) is given by \(\frac{-\mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{r}}\) The negative sign of potential energy reflects the stability of the electron in the ground state due to the coulombic force of attraction between the nucleus and electron. for \(\mathrm{Li}^{2+} \mathrm{Z}=3\), where \(\mathrm{Z}=\) atomic number of an atom Thus the potential energy of the electron present in the ground state of \(\mathrm{Li}+2\) ion is \(\frac{-3 e^2}{4 \pi \varepsilon_0 r}\) ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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