Question

The energy of the second stationary state in $Li^{+2}$ ion is $-4.95 \times 10^{-18}$ J. Find ionization energy (in KJ/mole) for $He^{+}$ ions in its ground state ($N_A = 6 \times 10^{23}$).

Answer 1

5280 KJ/mole

Answer 2

The energy of an electron in the n-th stationary state of a hydrogen-like atom is given by $E_n = -k \frac{Z^2}{n^2}$. For $Li^{+2}$ ($Z=3$) in the second stationary state ($n=2$), $E_2 = -4.95 \times 10^{-18}$ J. Using this, we find $k$: $-4.95 \times 10^{-18} = -k \frac{3^2}{2^2} \implies k = \frac{4.95 \times 10^{-18} \times 4}{9} = 2.2 \times 10^{-18}$ J. For $He^{+}$ ($Z=2$) in its ground state ($n=1$), the energy is $E_1 = -k \frac{Z^2}{n^2} = -(2.2 \times 10^{-18}) \frac{2^2}{1^2} = -8.8 \times 10^{-18}$ J. The ionization energy ($IE$) is the energy required to remove the electron from $n=1$ to $n=\infty$, so $IE = E_{\infty} - E_1 = 0 - E_1 = -E_1 = 8.8 \times 10^{-18}$ J per ion. To convert to KJ/mole: $IE (\text{KJ/mole}) = (8.8 \times 10^{-18} \text{ J/ion}) \times (6 \times 10^{23} \text{ ions/mole}) \times (\frac{1 \text{ KJ}}{1000 \text{ J}}) = 5280$ KJ/mole.