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A hydrogen-like ion has one electron and a nucleus of charge +Ze…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A hydrogen-like ion has one electron and a nucleus of charge $+Ze$ . When its electron falls from the orbit $n = 3$ to the orbit $n = 2$ , it gives out light of wavelength $72.94\ \text{nm}$ . For such an ion the wavenumber of a line is $\bar{\nu} = R_H Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ , the energy of orbit $n$ is $E_n = -2.18 \times 10^{-18}\,\dfrac{Z^2}{n^2}\ \text{J}$ , and its radius is $r_n = 52.9\,\dfrac{n^2}{Z}\ \text{pm}$ . Take $R_H = 109\,677\ \text{cm}^{-1}$ , $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ , $h = 6.626 \times 10^{-34}\ \text{J s}$ and $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$ .

Find $Z$ . Which ion is it?
What is the energy of the electron when the ion is in its ground state?
What is the ionisation energy of the ion from its ground state, in $\text{kJ mol}^{-1}$ ?
What is the radius of the orbit $n = 2$ ?
The lines that end on $n = 2$ form a series. What is the shortest wavelength in this series?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Use the one line to find $Z$ ; everything else follows from the Bohr formulas.

(a) The nuclear charge. $\bar{\nu} = \dfrac{1}{72.94 \times 10^{-7}\ \text{cm}} = 137099\ \text{cm}^{-1}$ , and $\dfrac14 - \dfrac19 = \dfrac{5}{36}$ . So

\displaystyle Z^2 = \frac{\bar{\nu}}{R_H \times \frac{5}{36}} = \frac{137099}{109\,677 \times 0.1389} = 9.00

So $Z = 3$ : the ion is $\text{Li}^{2+}$ .

(b) Ground state. $E_1 = -2.18 \times 10^{-18} \times 3^2\ \text{J} = -1.962 \times 10^{-17}\ \text{J}$ .

(c) Ionisation energy. For one ion it is $-1.962 \times 10^{-17}$ with the sign changed. For a mole:

\displaystyle 1.962 \times 10^{-17}\ \text{J} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 11815\ \text{kJ mol}^{-1} \approx 1.18 \times 10^{4}\ \text{kJ mol}^{-1}

(d) Orbit $n = 2$ . $r_2 = 52.9 \times \dfrac{2^2}{3}\ \text{pm} = 70.5\ \text{pm}$ .

(e) Series limit. With $n_1 = 2$ and $n_2 = \infty$ : $\bar{\nu} = 109\,677 \times \dfrac{9}{4}\ \text{cm}^{-1}$ , so

\displaystyle \lambda = \frac{4}{109\,677 \times 9}\ \text{cm} = 40.52\ \text{nm}

Check. From the energies: the electron in $n = 2$ has $|E_2| = 4.905 \times 10^{-18}\ \text{J}$ , and $\lambda = hc/|E_2| = 40.53\ \text{nm}$ , the same.

Watch out. $Z$ comes in squared in the energy and the wavenumber, but only to the first power in the radius.

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