Particles Behaving as Waves Problems and Solutions 13

Q#39.91 

Structure of a Virus. To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. 

(a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? 

(b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

Answer:

The wave (light or electron matter wave) having less energy will cause less damage to the virus.

For a photon $E_{ph}=\frac{hc}{\lambda}$ and

For an electron $E_e=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}$

(a) $E_{ph}=\frac{(6.626 \times 10^{-34} \ J.s)(2.998 \times 10^8 \ m/s)}{5.00 \times 10^{-9} \ m}$

$E_{ph}=248 \ eV$

(b) $E_e=\frac{(6.626 \times 10^{-34} \ J.s)^2}{2(9.11 \times 10^{-31} \ kg)(5.00 \times 10^{-9} \ m)^2}$

$E_e = 9.65 \times 10^{-21}$ J

$E_e = \frac{9.65 \times 10^{-21} \ J}{1.6 \times 10^{-19} \ J/eV}$ 

$E_e=0.0603 \ eV$

The electron has much less energy than a photon of the same wavelength and therefore would cause much less damage to the virus. 

Q#39.92

Zero-Point Energy. Consider a particle with mass m moving in a potential $U = \frac{1}{2}kx^2$, as in a mass–spring system. The total energy of the particle is E = $p^2/2m + \frac{1}{2}kx^2$. Assume that p and x are approximately related by the Heisenberg uncertainty principle, so px ≈ h. 

(a) Calculate the minimum possible value of the energy E, and the value of x that gives this minimum E. This lowest possible energy, which is not zero, is called the zero-point energy. 

(b) For the x calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle?

Answer:

Assume px ≈ h and use this to express E as a function of x. E is a minimum for that x that satisfies $\frac{dE}{dx}=0$

(a) Using the given approximation,

E = $\frac{1}{2}\left(\frac{(h/x)^2}{m}+kx^2\right)$  

E = $\frac{1}{2}\left(\frac{h^2}{mx^2}+kx^2\right)$  (*)

E = $\frac{1}{2}\left(\frac{h^2}{m}x^{-2}+kx^2\right)$  

$\frac{dE}{dx}$ = $\frac{1}{2}\left(-\frac{h^2}{2m}x^{-3}+2kx\right)$  

$\frac{dE}{dx}$ = $-\frac{h^2}{mx^3}+kx$  

and the minimum energy occurs when $\frac{dE}{dx}=0$, so

kx = $\frac{h^2}{mx^3}$ or

$x^4=\frac{h^2}{mk}$

$x^2=\frac{h}{\sqrt{mk}}$

The minimum energy occurs at

$x=\left(\frac{h^2}{mk}\right)^{1/4}$

We substitute the value of $x^2$ in (*) to get the minimum energy

E = $\frac{1}{2}\left(\frac{h^2}{m\frac{h}{\sqrt{mk}}}+k\frac{h}{\sqrt{mk}}\right)$

= $\frac{1}{2}\left(\frac{h\sqrt{mk}}{m}+\frac{kh}{\sqrt{mk}}\right)$

= $\frac{1}{2}\left(\frac{hk}{\sqrt{mk}}+\frac{kh}{\sqrt{mk}}\right)$

= $\frac{hk}{\sqrt{mk}}$

= $\frac{hk}{\sqrt{\frac{mk^2}{k}}}$

= $\frac{hk}{\sqrt{k\frac{m}{k}}}$

E = $h\sqrt{\frac{k}{m}}$

(b) $U=\frac{1}{2}kx^2$

$U=\frac{1}{2}k\left(\frac{h}{\sqrt{mk}}\right)$

$=\frac{1}{2}\left(\frac{hk}{\sqrt{\frac{mk^2}{k}}}\right)$

$=\frac{1}{2}\left(\frac{hk}{k\sqrt{\frac{m}{k}}}\right)$

$=\frac{1}{2}\left(\frac{h}{\sqrt{\frac{m}{k}}}\right)$

$U=\frac{h}{2}\sqrt{\frac{k}{m}}$

and kinetic energy $K=\frac{p^2}{2m}=\frac{h^2}{2mx^2}$

$K=\frac{h^2}{2m\frac{h}{\sqrt{mk}}}=\frac{h\sqrt{mk}}{2m}$

$K=\frac{h}{2}\sqrt{\frac{k}{m}}$

At this x the kinetic and potential energies are the same.

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