6.5 De broglie’s matter waves  (Page 2/10)

Depending on the problem at hand, in this equation we can use the following values for hc : $hc=(6.626\times {10}^{\mathrm{-34}}\text{J}·\text{s})(2.998\times {10}^8\text{m/s})=1.986\times {10}^{\mathrm{-25}}\text{J}·\text{m}=1.241\text{eV}·\mu \text{m}$

Solution

1. For the basketball, the kinetic energy is
   
   $K=m_0{u}^2\text{/}2=(0.65\text{kg}){(10\text{m/s})}^2\text{/}2=32.5\text{J}$
   
   and the rest mass energy is
   
   $E_0=m_0{c}^2=(0.65\text{kg}){(2.998\times {10}^8\text{m/s})}^2=5.84\times {10}^{16}\text{J.}$
   
   We see that $K\text{/}(K+E_0)\ll 1\text{and use}p=m_{0}u=(0.65\text{kg})(10\text{m/s})=6.5\text{J}·\text{s/m}:$
   
   $\lambda =\frac{h}{p}=\frac{6.626\times {10}^{\mathrm{-34}}\text{J}·\text{s}}{6.5\text{J}·\text{s/m}}=1.02\times {10}^{\mathrm{-34}}\text{m}.$
2. For the nonrelativistic electron,
   
   $E_0=m_0{c}^2=(9.109\times {10}^{\mathrm{-31}}\text{kg}){(2.998\times {10}^8\text{m/s})}^2=511\text{keV}$
   
   and when $K=1.0\text{eV},$ we have $K\text{/}(K+E_0)=(1\text{/}512)\times {10}^{\mathrm{-3}}\ll 1,$ so we can use the nonrelativistic formula. However, it is simpler here to use [link] :
   
   $\lambda =\frac{h}{p}=\frac{hc}{\sqrt{K(K+2E_0)}}=\frac{1.241\text{eV}·\mu \text{m}}{\sqrt{(1.0\text{eV})[1.0\text{eV+}2(511\text{keV})]}}=1.23\text{nm}.$
   
   If we use nonrelativistic momentum, we obtain the same result because 1 eV is much smaller than the rest mass of the electron.
3. For a fast electron with $K=108\text{keV,}$ relativistic effects cannot be neglected because its total energy is $E=K+E_0=108\text{keV}+511\text{keV}=619\text{keV}$ and $K\text{/}E=108\text{/}619$ is not negligible:
   
   $\lambda =\frac{h}{p}=\frac{hc}{\sqrt{K(K+2E_0)}}=\frac{1.241\text{eV}·\text{μm}}{\sqrt{108\text{keV}[108\text{keV}+2(511\text{keV})]}}=3.55\text{pm}.$

Significance

Using the concept of the electron matter wave, de Broglie provided a rationale for the quantization of the electron’s angular momentum in the hydrogen atom, which was postulated in Bohr’s quantum theory. The physical explanation for the first Bohr quantization condition comes naturally when we assume that an electron in a hydrogen atom behaves not like a particle but like a wave. To see it clearly, imagine a stretched guitar string that is clamped at both ends and vibrates in one of its normal modes. If the length of the string is l ( [link] ), the wavelengths of these vibrations cannot be arbitrary but must be such that an integer k number of half-wavelengths $\lambda \text{/}2$ fit exactly on the distance l between the ends. This is the condition $l=k\lambda \text{/}2$ for a standing wave on a string. Now suppose that instead of having the string clamped at the walls, we bend its length into a circle and fasten its ends to each other. This produces a circular string that vibrates in normal modes, satisfying the same standing-wave condition, but the number of half-wavelengths must now be an even number $k,k=2n,$ and the length l is now connected to the radius $r_n$ of the circle. This means that the radii are not arbitrary but must satisfy the following standing-wave condition:

If an electron in the n th Bohr orbit moves as a wave, by [link] its wavelength must be equal to $\lambda =2\pi r_n\text{/}n.$ Assuming that [link] is valid, the electron wave of this wavelength corresponds to the electron’s linear momentum, $p=h\text{/}\lambda =nh\text{/}(2\pi r_n)=n\hslash \text{/}{r}_n.$ In a circular orbit, therefore, the electron’s angular momentum must be

This equation is the first of Bohr’s quantization conditions, given by [link] . Providing a physical explanation for Bohr’s quantization condition is a convincing theoretical argument for the existence of matter waves.