A gas of ions, each with a single electron, is excited and glows in the ultraviolet…

A gas of ions, each with a single electron, is excited and glows in the ultraviolet. In its spectrum one series has all its lines ending on the same lower level of the ion, whose quantum number $m$ is not known. The figure shows the two longest wavelengths of that series, $\lambda_1$ and $\lambda_2$ , with the shorter lines crowding together beyond them. Use the Rydberg constant $R = 1.097 \times 10^{7}\ \text{m}^{-1}$ , the ground state energy of hydrogen $13.6\ \text{eV}$ and the first Bohr radius $52.9\ \text{pm}$ . The nucleus carries a charge $+Ze$ , where $Z$ is not known either. The ions are all of one kind, and each has lost every electron but one, so the Bohr model for a single electron applies.

A strip representing an ultraviolet emission spectrum of a hydrogen-like ion, with a wavelength scale in nanometres from 80 to 180. Two dark lines stand on the strip, one at 121.5 nm and one at 164.1 nm. Beyond the line at 121.5 nm to the left, a few fainter lines are drawn crowding towards a short wavelength end. A note says that the two marked lines are the two longest wavelengths of one series.

What is the quantum number $m$ of the lower level?
What is the atomic number $Z$ of the ion?
What is the shortest wavelength of this series, the series limit, in nanometres?
What energy is needed to remove the electron from the ion in its ground state, in electron volts?
What is the radius of the orbit with quantum number $m$ , in picometres?

The ratio $\lambda_1 / \lambda_2$ depends on $m$ only, so test it first.

The worked solution

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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.

(a) The lower level. The longest line is the jump $m + 1 \to m$ and the next is $m + 2 \to m$ . So $\dfrac{\lambda_1}{\lambda_2} = \dfrac{\frac{1}{m^2} - \frac{1}{(m+2)^2}}{\frac{1}{m^2} - \frac{1}{(m+1)^2}}$ , and $Z$ cancels. From the figure, $\dfrac{164.1}{121.5} = 1.351$ . The formula gives $1.185$ for $m = 1$ , $1.350$ for $m = 2$ and $1.463$ for $m = 3$ . So $m = 2$ .

(b) The atomic number. For $m = 2$ , $\dfrac{1}{\lambda_1} = RZ^2\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = \dfrac{5RZ^2}{36}$ . So $Z^2 = \dfrac{36}{5R\lambda_1}$ . With $R = 1.097 \times 10^{7}\ \text{m}^{-1}$ and $\lambda_1 = 164.1\ \text{nm}$ , $Z^2 = 4.00$ , and $Z = 2$ . The ion is $\text{He}^+$ .

(c) The series limit. The jump from $n = \infty$ to $m = 2$ gives $\dfrac{1}{\lambda} = \dfrac{RZ^2}{4}$ . So $\lambda = \dfrac{4}{1.097 \times 10^{7}\ \text{m}^{-1} \times 4} = 91.2\ \text{nm}$ .

(d) Ionisation energy. The ground state energy is $-13.6\ \text{eV} \times Z^2 = -54.4\ \text{eV}$ , so $54.4\ \text{eV}$ is needed to free the electron.

(e) The orbit. $r_n = \dfrac{n^2}{Z} \times 52.9\ \text{pm}$ . For $n = 2$ and $Z = 2$ , $r = \dfrac{4}{2} \times 52.9\ \text{pm} = 105.8\ \text{pm}$ .

Answer: (a) $2$ . (b) $2$ . (c) $91.2\ \text{nm}$ . (d) $54.4\ \text{eV}$ . (e) $105.8\ \text{pm}$ .

Check. The energy of the longest line is $7.56\ \text{eV}$ for $Z = 2$ between $n = 3$ and $n = 2$ , and $\dfrac{1240\ \text{eV nm}}{7.56\ \text{eV}} = 164.1\ \text{nm}$ , the wavelength in the figure. The limit corresponds to $\dfrac{13.6 \times 4}{4}\ \text{eV} = 13.6\ \text{eV}$ , and $\dfrac{1240\ \text{eV nm}}{13.6\ \text{eV}} = 91.2\ \text{nm}$ .

Watch out. One wavelength alone fits any $m$ if $Z$ is allowed to take an odd value, so it cannot fix both. The ratio of two lines fixes $m$ first, and only then does $Z$ come out as a whole number.

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