Question 42.4: Tungsten Although Bohr’s model doesn’t work well for complic......
Tungsten
Although Bohr’s model doesn’t work well for complicated atoms, it does fit ions that have been stripped of all but one electron. Consider such an ion of tungsten. If the ion goes from its first excited state (n = 2) down to the ground state, what is the energy of the photon emitted?
Interpret and Anticipate
We start by modifying the expression (Eq. 42.9)
for hydrogen’s energy levels to find an expression for ionized tungsten’s energy levels. Then we can find the energy between the first excited state and the ground state.
Solve
Revisit the derivation of the hydrogen spectral lines (Eq. 42.3).
We started by applying Coulomb’s law to the single proton and electron in a hydrogen atom. When we apply Coulomb’s law to ionized tungsten, we must take into account that it has Z protons, where Z is its atomic number. So the speed of the electron (Eq. 1 in the derivation) increases by a factor of Z.
For hydrogen
F_E=k \frac{q_1 q_2}{r^2}=k \frac{e^2}{r^2}For other ions (with one electron) and Z protons
\begin{aligned}& F_E=k \frac{q_1 q_2}{r^2}=k \frac{Z e^2}{r^2} \\& \nu^2=\frac{k Z e^2}{m_e r} \quad \quad (1)\end{aligned}This higher speed means that the kinetic energy is higher by a factor of Z.
For hydrogen
K=\frac{1}{2} m_e \nu^2=\frac{1}{2} \frac{k e^2}{r}For other ions (with one electron) and Z protons
K=\frac{1}{2} \frac{k Z e^2}{r} \quad \quad (2)Continue to revisit the derivation, and you find that the radius of the electron’s orbit (Eq. 4 in the derivation) is reduced by this factor of Z.
For hydrogen
r=\frac{n^2 h^2}{4 \pi^2 k m_e e^2}For other ions (with one electron) and Z protons
r=\frac{n^2 h^2}{4 \pi^2 k m_e Z e^2} \quad \quad (3)There is another modification we must make to the derivation. The potential energy of the nucleus-electron system is modified by a factor of Z.
For hydrogen
U_E=k \frac{Q q}{r}=-\frac{k e^2}{r}For other ions (with one electron) and Z protons
U_E=-\frac{k Z e^2}{r} \quad \quad (4)Combine equations (2) through (4) to find an expression for the energy levels in an ion with only one electron remaining.
\begin{aligned}& E=K+U=-\frac{1}{2} \frac{k Z e^2}{r} \\& E=-\frac{1}{2} k Z e^2\left(\frac{4 \pi^2 k m_e Z e^2}{n^2 h^2}\right)=-\frac{2 \pi^2 k^2 m_e Z^2 e^4}{h^2} \frac{1}{n^2}\end{aligned}Compare the expression here to Equation 42.9 for hydrogen’s energy levels, and you see that they differ by a factor of Z².
For hydrogen
E=-\frac{2 \pi^2 k^2 m_e e^4}{h^2} \frac{1}{n^2} \quad \quad (42.9)We find a convenient expression for the energy levels in the ion by multiplying Equation 42.10 by Z².
E_n=-\frac{13.6 eV }{n^2} \quad \quad (42.10)For other ions (with one electron) and Z protons
E_{ ion }=Z^2 E_{\text {Hydrogen }}=-\frac{Z^2(13.6 eV )}{n^2} \quad \quad (5)From the periodic table (Appendix B), tungsten’s atomic number is 74, which we substitute in Equation (5).
E_{\text {Tungsten }}=-\frac{74^2(13.6 eV )}{n^2}=-\frac{74.5 keV }{n^2} \quad \quad (6)We use Equation (6) to write an expression for the difference in the tungsten ion’s energy levels, which corresponds to Equation 4 in the derivation for the hydrogen spectral lines. Then we substitute for the energy levels. The photon carries away energy equal to the difference in these levels.
\begin{aligned}& \Delta E_{\text {Tungsten }}=74.5 keV \left(\frac{1}{m^2}-\frac{1}{n^2}\right) \\& \Delta E_{\text {Tungsten }}=74.5 keV \left(\frac{1}{1^2}-\frac{1}{2^2}\right)=55.9 keV \\& E_{\text {photon }}=55.9 keV\end{aligned}Check and Think
When tungsten is stripped of all but one electron, it has 74 protons and 1 electron. We expect that when the ion is in its ground state, it should be much more tightly bound than when a hydrogen atom is in its ground state. This is exactly what we find when we compare Equation (6) in this example for the tungsten ion to Equation 42.10 for hydrogen. The numerator in these equations indicates the amount of energy it would take to free the electron from its ground state. In the case of hydrogen, this is 13.6 eV. For tungsten the energy requirement is roughly 74500 eV—a factor of 74² = 5476 times greater.
