Sunday, February 20, 2011

Mastering Physics: De Broglie Waves in the Bohr Model

The hypothesis that was put forward by Louis de Broglie in 1924 was astonishing for a number of reasons. An obvious reason is that associating a wavelike nature with particles is far from intuitive, but another astonishing aspect was how well the hypothesis fit in with certain parts of existing physics. In this problem, we explore the correspondence between the de Broglie picture of the wave nature of electrons and the Bohr model of the hydrogen atom.

Part A
What is the de Broglie wavelength lambda_1 of the electron in the first Bohr energy level of the hydrogen atom?
Express your answer in terms of epsilon_0, h, the mass of the electron m, and the magnitude of the charge on the electron e.
ANSWER:

  lambda  = 2h^{2}\frac{{\epsilon}_{0}}{me^{2}}

Part B
What is the cirumference c_1 of the first (n=1) energy level in hydrogen predicted by the Bohr model of the atom?
Express your answer in terms of epsilon_0, h, m, and e.
ANSWER:

  c_1  = \frac{2h^{2}{\epsilon}_{0}}{me^{2}}

Part C
What is the de Broglie wavelength lambda_3 of the electron in the third (n=3) Bohr energy level of the hydrogen atom?
Express your answer in terms of epsilon_0, h, m, and e.
ANSWER:

  lambda_3  = \frac{6h^{2}{\epsilon}_{0}}{me^{2}}

Part D
What is the circumference c_3 of the third (n=3) energy level in hydrogen predicted by the Bohr model of the atom?
Express your answer in terms of epsilon_0, h, m, and e.
ANSWER:

  c_3  = \frac{18h^{2}{\epsilon}_{0}}{me^{2}}
 In contrast to what we saw at the first energy level, the de Broglie wavelength of the electron in the third energy level is not equal to the circumference of the electron's orbit as predicted by the Bohr model. In this case, it is equal to one-third of the circumference of the orbit.


Part E
In the previous parts, you saw that there is not equality between the de Broglie wavelength of an electron in the hydrogen atom and the circumference of its orbit. However, there does exist a definite relationship. What is the relationship between the circumference of the orbit of the n th energy level c_n and the de Broglie wavelength lambda_n?
Express your answer in terms of n and lambda_n.
ANSWER:

  c_n  = n{\lambda}_{n}
As you can now see, the relationship between the de Broglie wavelength of an electron and the circumference of its orbit is that the circumference is equal to n \lambda_n. An interesting interpretation of this result is that the n th energy level has n nodes. Although quantum physics soon developed more sophisticated modes of analysis than those of de Broglie and Bohr, the relationship between nodes in a wave and the energy state is something that you will see again.
Posted by Gymnos at 7:31 PM
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1 comment:

  1. UnknownNovember 20, 2012 at 10:02 AM

    How do we know that the circumference of the electron shell is the same as one wavelength?

    ReplyDelete

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