Problem: X-rays from ionised fluorine

Problem

A fluorine atom ($Z = 9$) loses one of its innermost electrons, and another electron drops from level $n = 3$ to level $n = 2$ to fill the vacancy. Given that, for hydrogen-like ions, the levels scale as $E_n^{(Z)} = -Z^2\cdot 13{,}6/n^2$ eV, calculate the energy of the emitted photon and its wavelength.

Solution

Step 1 — Hydrogen-like levels. For an ion with a single effective charge $Z$, the levels are those of hydrogen multiplied by $Z^2$. With $Z = 9$ we get $Z^2 = 81$.

$E_3 = -\frac{81\cdot 13{,}6}{3^2} = -\frac{81\cdot 13{,}6}{9} = -122{,}4\;\text{eV}$

$E_2 = -\frac{81\cdot 13{,}6}{2^2} = -\frac{81\cdot 13{,}6}{4} = -275{,}4\;\text{eV}$

Step 2 — Photon energy. In the $n=3\to n=2$ transition the emitted photon carries away the energy difference:

$\Delta E = E_3 - E_2 = -122{,}4 - (-275{,}4) = 153\;\text{eV}$

Step 3 — Wavelength. Using the rule of thumb $\lambda(\text{nm}) = 1240/E(\text{eV})$:

$\lambda = \frac{1240}{153} \approx 8{,}1\;\text{nm}$

A wavelength of a few nanometres falls in the soft X-ray band, consistent with the name of the phenomenon.

$\ev{\Delta E \approx 153\;\text{eV}, \qquad \lambda \approx 8\;\text{nm}}$

Links

Topics: Quantum physics Concepts: Bohr model · Photon Skills: Symbolic setup · Unit conversion