Particles Behaving as Waves Problems and Solutions 12

Q#39.88

For x rays with wavelength 0.0300 nm, the m = 1 intensity maximum for a crystal occurs when the angle $\theta$ in Fig. 39.2 is At $35.8^0$ what angle $\theta$ does the m = 1 maximum occur when a beam of 4.50-keV electrons is used instead? Assume that the electrons also scatter from the atoms in the surface plane of this same crystal.

Answer:

Assume both the x rays and electrons are at normal incidence and scatter from the surface plane of the crystal, so the maxima are located by sin , d sin θ = mλ where d is the separation between adjacent atoms in the surface plane. 

Let primed variables refer to the electrons.

$\lambda'=\frac{h}{p'}=\frac{h}{\sqrt{2mE'}}$

$sin \ \theta' = \frac{\lambda'}{\lambda} sin \ \theta$

$sin \ \theta' = \frac{h}{\lambda \sqrt{2mE'}} sin \ \theta$

so, $\theta' = arcsin \frac{h}{\lambda \sqrt{2mE'}} sin \ \theta$

$\theta' = arcsin \frac{(6.626 \times 10^{-34} \ J.s)(sin \ 35.8^0)}{(3.00 \times 10^{-11} \ m) \sqrt{2(9.11 \times 10^{-31} \ kg)(4.50 \times 10^3 \ eV)(1.6 \times 10^{-19 J/eV})}}$

$\theta'=20.9^0$

The x rays and electrons have different wavelengths and the 1 m = maxima occur at different angles.

Q#39.89 

Electron diffraction can also take place when there is interference between electron waves that scatter from atoms on the surface of a crystal and waves that scatter from atoms in the next plane below the surface, a distance d from the surface (see Fig. 36.23c). 

(a) Find an equation for the angles $\theta$ at which there is an intensity maximum for electron waves of wavelength $\lambda$.

(b) The spacing between crystal planes in a certain metal is 0.091 nm. If 71.0-eV electrons are used, find the angle at which there is an intensity maximum due to interference between scattered waves from adjacent crystal planes. The angle is measured as shown in Fig. 36.23c. 

(c) The actual angle of the intensity maximum is slightly different from your result in part (b). The reason is the work function $\phi$ of the metal (see Section 38.1), which changes the electron potential energy by $-e \phi$ when it moves from vacuum into the metal. If the effect of the work function is taken into account, is the angle of the intensity maximum larger or smaller than the value found in part (b)? Explain.

Answer:

The interference pattern for electrons with de Broglie wavelength λ is the same as for light with wavelength λ.

For an electron, $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}$ 

(a) $\lambda =\frac{6.626 \times 10^{-34} \ J.s}{\sqrt{2(9.11 \times 10^{-31} \ kg)(71.0 \ eV)}(1.6 \times 10^{-19} \ J/eV)}$

$\lambda = 1.46 \times 10^{-10}$ m

The maxima occur when d sin θ = m$\lambda$

sin θ = $\left(\frac{m\lambda}{2d}\right)$

Note: This m is the order of the maximum, not the mass.

sin θ = $\left(\frac{(1)(1.46 \times 10^{-10} \ m)}{2(9.10 \times 10^{-11} \ m)}\right)$

θ = $sin^{-1}(0.8022)=53.3^0$

(c) The work function of the metal acts like an attractive potential increasing the kinetic energy of incoming electrons by $e\phi$. An increase in kinetic energy is an increase in momentum that leads to a smaller wavelength. A smaller wavelength gives a smaller angle θ (see part (b)). 

Q#39.90

A certain atom has an energy level 3.50 eV above the ground state. When excited to this state, it remains 4.0 $\mu$m on the average, before emitting a photon and returning to the ground state. 

(a) What is the energy of the photon? What is its wavelength? 

(b) What is the smallest possible uncertainty in energy of the photon?

Answer:

The photon is emitted as the atom returns to the lower energy state. The duration of the excited state limits the energy of that state due to the uncertainty principle.

The wavelength λ of the photon is related to the transition energy E of the atom by E = $\frac{hc}{\lambda}$

 From $\Delta E \Delta t \geq \frac{\hbar}{2}$, the minimum uncertainty in energy is

$\Delta E  \geq \frac{\hbar}{2\Delta t}$

(a) The photon energy equals the transition energy of the atom, 3.50 eV. 

$\lambda$ = $\frac{hc}{E}$

$\lambda$ = $\frac{(4.136 \times 10^{-15} \ eV.s)(2.998 \times 10^8 \ m/s)}{3.50 \ eV}$

$\lambda = 355 \ nm$

(b) $\Delta E  = \frac{\hbar}{2\Delta t}$

$\Delta E  = \frac{1.055 \times 10^{-34} \ J.s}{2(4.0 \times 10^{-6} \ s)}$

$\Delta E = 1.32 \times 10^{-29} \ J = 8.12 \times 10^{-11} \ eV$

 The uncertainty in the energy could be larger than that found in (b), but never smaller.

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