Particles Behaving as Waves Problems and Solutions 6

Q#39.70

Electrons go through a single slit 150 nm wide and strike a screen 24.0 cm away. You find that at angles of $\pm 20.0^0$ from the center of the diffraction pattern, no electrons hit the screen but electrons hit at all points closer to the center. (a) How fast were these electrons moving when they went through the slit? (b) What will be the next larger angles at which no electrons hit the screen? 

Answer:

The de Broglie wavelength of the electrons must be such that the first diffraction minimum occurs at θ = 20.0°.

The single-slit diffraction minima occur at angles θ given by a sin θ = mλ. 

(a) $\lambda = a sin \theta = (150 \times 10^{-9} \ m)sin 20.0^0 = 5.13 \times 10^{-8} \ m$

For a particle with mass, $p = \frac{h}{\lambda}$, so 

$\lambda = \frac{h}{p}=\frac{h}{mv}$

$v = \frac{h}{m\lambda}=\frac{6.626 \times 10^{-34} \ J.s}{(9.11 \times 10^{-31} \ kg)(5.13 \times 10^{-8} \ m)}$

$v = 1.42 \times 10^4 \ m/s$

(b) No electrons strike the screen at the location of the second diffraction minimum.

$a sin\theta_2=2\lambda$

$sin\theta_2= \pm 2\frac{\lambda}{a}$ 

$sin\theta_2= \pm 2\frac{5.13 \times 10^{-8} \ m}{150 \times 10^{-9} \ m}=\pm 0.684$ 

$\theta_2=\pm 43.2^0$

 The intensity distribution in the diffraction pattern depends on the wavelength λ and is the same for light of wavelength λ as for electrons with de Broglie wavelength .

Q#39.71

A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 1.25 nm apart. You observe that the first double-slit interference dark fringe occurs at $\pm 18.0^0$ from the original direction of the beam when viewed on a distant screen. (a) Are these electrons relativistic? How do you know? (b) Through what potential difference were the electrons accelerated?

Answer:

The electrons behave like waves and produce a double-slit interference pattern after passing through the slits.

The first angle at which destructive interference occurs is given by $d \ sin \ \theta = \frac{\lambda}{2}$.

The de Broglie wavelength of each of the electrons is $\lambda = \frac{h}{mv}$.

(a) First find the wavelength of the electrons. For the first dark fringe, we have $d \ sin \ \theta = \frac{\lambda}{2}$, which gives 

$(1.25 \ nm)(sin \ 18.0^0)=\frac{\lambda}{2}$

$\lambda = 0.7725 \ nm$

Now solve the de Broglie wavelength equation for the speed of the electron:

$v = \frac{h}{mv}$

$v = \frac{6.626 \times 10^{-34} \ J.s}{(9.11 \times 10^{-31} \ kg)(0.7725 \times 10^{-9} \ m)}=9.42 \times 10^5 \ m/s$

which is about 0.3% the speed of light, so they are nonrelativistic.

(b) Energy conservation gives $eV=\frac{1}{2}mv^2 $ and  

$V= \frac{mv^2}{2e}$

$V= \frac{(9.11 \times 10^{-31} \ kg)(9.42 \times 10^5 \ m/s)^2}{2(1.6 \times 10^{-19} \ C)}$

V = 2.52 volt

The hole must be much smaller than the wavelength of visible light for the electrons to show diffraction. 

Q#39.72

A beam of protons and a beam of alpha particles (of mass $6.64 \times 10^{-27}$ kg and charge +2e) are accelerated from rest through the same potential difference and pass through identical circular holes in a very thin, opaque film. When viewed far from the hole, the diffracted proton beam forms its first dark ring at  $\pm 15^0$  with respect to its original direction. When viewed similarly, at what angle will the alpha particle form its first dark ring?

Answer:

The alpha particles and protons behave as waves and exhibit circular-aperture diffraction after passing through the hole. 

 For a round hole, the first dark ring occurs at the angle θ for which sin θ = $\frac{1.22\lambda}{D}$ where D is the diameter of the hole. The de Broglie wavelength for a particle is $\lambda = \frac{h}{p}=\frac{h}{mv}$

Taking the ratio of the sines for the alpha particle and proton gives

$\frac{sin \theta_{\alpha}}{sin \theta_p}=\frac{1.22 \lambda_{alpha}}{1.22\lambda_p}=\frac{\lambda_{\alpha}}{\lambda_p}$

The de Broglie wavelength gives $\lambda_p=\frac{h}{p_p}$ and $\lambda_{\alpha}=\frac{h}{p_{\alpha}}$, so

$\frac{sin \theta_{\alpha}}{sin \theta_p}=\frac{h/p_{\alpha}}{h/p_p}=\frac{p_p}{p_{\alpha}}$

Using $K = \frac{p^2}{2m}$ we have $p=\sqrt{2mK}$.

Since the alpha particle has twice the charge of the proton and both are accelerated through the same potential difference, $K_{\alpha}=2K_p$.  Therefore

$p_p=\sqrt{2m_pK_p}$ and

$p_{\alpha}=\sqrt{2m_{\alpha}K_{\alpha}}$  

$p_{\alpha}=\sqrt{2m_{\alpha}(2K_p)}=\sqrt{4m_{\alpha}K_p}$

Substituting these quantities into the ratio of the sines gives 

$\frac{sin \theta_{\alpha}}{sin \theta_p}=\frac{p_p}{p_{\alpha}}$

$\frac{sin \theta_{\alpha}}{sin \theta_p}=\frac{\sqrt{2m_pK_p}}{\sqrt{4m_{\alpha}K_p}}$

$\frac{sin \theta_{\alpha}}{sin \theta_p}=\sqrt{\frac{m_p}{2m_{\alpha}}}$

Solving for sin $\theta_{\alpha}$ gives

$sin \ \theta_{\alpha}=\sqrt{\frac{m_p}{2m_{\alpha}}}sin \ \theta_p$

$sin \ \theta_{\alpha}=\sqrt{\frac{1.67 \times 10^{-27} \ kg}{2(6.64 \times 10^{-27} \ kg)}}sin \ 15.0^0$

$sin \ \theta_{\alpha}=0.0918 $

$\theta_{\alpha}=5.27^0 $

Since sin θ is inversely proportional to the mass of the particle, the larger-mass alpha particles form their first dark ring at a smaller angle than the ring for the lighter protons.

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