Sunday, February 13, 2011

Mastering Physics: Hydrogen-Like Atoms

According to the Bohr model of a hydrogen atom, the frequency of light radiated by an electron moving from an orbit n_1 to an orbit n_2 corresponds to the energy level difference between n_1 and n_2 of
E = E_{0}\left(\frac{1}{n_{1}^2}-\frac{1}{n_{2}^2}\right),
where
E_{0} = -\frac{m_{e}Z^{2}e^{4}}{32\pi^{2}\epsilon_{0}^{2}\hbar^{2}},
and where m_e is the electron mass, Z is the atomic number, e is the magnitude of the electron charge, epsilon_0 is the permittivity of free space, and hbar is Planck's constnt divided by 2\pi. In the case of hydrogen (Z = 1) E_{0}=-13.6\; {\rm eV}.

Part A
Find the frequency of light f radiated by an electron moving from orbit n_{1}=2 to n_{2}=1 inside of a {\rm He}^+ ion.
Express your answer in hertz to three significant figures.
ANSWER:

  f = 9.860×1015
 {\rm Hz}


Part B
In the Bohr model of hydrogen, the radius of the n^{\rm th} orbit is defined as
r_{n}= a_{0} \frac{n^{2}}{Z}\; ,
where
a_{0} = \frac{4\pi \epsilon \hbar^{2}}{m_{e}e^{2}} = 5.29 \times 10^{-11}\;{\rm m}
is called the Bohr radius. Find the radius r_1 of a valence orbital for a {\rm He}^+ ion.
Express your answer in meters to three significant figures.
ANSWER:

  r_1 = 2.65×10−11
 {\rm m}
 In fact, the radius of the helium valence orbital is 3.01 \times 10^{-11}\; {\rm m}. The discrepancy between your answer and the real value can be explained by the fact that the Bohr model is a semiclassical treatment of a quantum problem. It only works well for the hydrogen atom and, with some corrections, for helium. Bohr's model fails for larger atoms. This is due to the presence of many (more than one) electrons in those atoms. With a large number of electrons in an atom, their mutual interactions become significant and cannot be ignored, as they are in the Bohr model.

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