Q#39.73
An electron beam and a photon beam pass through identical slits. On a distant screen, the first dark fringe occurs at the same angle for both of the beams. The electron speeds are muchslower than that of light.
(a) Express the energy of a photon in terms of the kinetic energy K of one of the electrons.
(b) Which is greater, the energy of a photon or the kinetic energy of an electron?
Answer:
Both the electrons and photons behave like waves and exhibit single-slit diffraction after passing through their respective slits.
The energy of the photon is E = hc/λ and the de Broglie wavelength of the electron is λ = h/p = h/mv. Destructive interference for a single slit first occurs when a sin θ = λ.
(a) For the photon: λ = hc/E and a sin θ = λ. Since the a and θ are the same for the photons and electrons, they must both have the same wavelength. Equating these two expressions for λ gives
$a \ sin \ \theta=\frac{hc}{E}$
For the electron, $\lambda = \frac{h}{p}=\frac{h}{\sqrt{2mK}}$ and a sin θ = λ.
Equating these two expressions for λ gives
$a \ sin \ \theta = \frac{h}{p}=\frac{h}{\sqrt{2mK}}$
Equating the two expressions for a sin θ gives
$\frac{hc}{E} = \frac{h}{p}=\frac{h}{\sqrt{2mK}}$, which gives
$E = c\sqrt{2mK}$
(b) $\frac{E}{K}=\frac{c\sqrt{2mK}}{K}=\sqrt{\frac{2mc^2}{K}}$
Since v<<c, $mc^2$ > K, so the square root is > 1. Therefore E/K > 1, meaning that the photon has more energy than the electron.
When a photon and a particle have the same wavelength, the photon has more energy than the particle.
Q#39.74
Coherent light is passed through two narrow slits whose separation is The second-order bright fringe in the interference pattern is located at an angle of 0.0300 rad. If electrons are used instead of light, what must the kinetic energy (in electron volts) of the electrons be if they are to produce an interference pattern for which the second-order maximum is also at 0.0300 rad?
Answer:
The de Broglie wavelength of the electrons must equal the wavelength of the light.
The maxima in the two-slit interference pattern are located by d sin θ = mλ .
For an electron $\lambda=\frac{h}{p}=\frac{h}{mv}$
$\lambda=\frac{d \ sin \ \theta}{m}= \frac{(40.0 \times 10^{-6} \ m) \ sin \ 0.0300 \ rad}{2}=600 nm$.
The velocity of an electron with this wavelength is given by
$v = \frac{p}{m}=\frac{h}{m\lambda}$
$v = \frac{6.63 \times 10^{-34} \ J.s}{(2)(600 \times 10^{-9} \ m)} = 1.21 \times 10^3 \ m/s$
Since this velocity is much smaller than c we can calculate the energy of the electron classically
$K = \frac{1}{2}mv^2 = \frac{1}{2}(9.11 \times 10^{-31} \ kg)(1.21 \times 10^3 \ m/s)^2$
$K = 6.70 \times 10^{-25} \ J=4.19 \times 10^{-6} \ m = 4.19 \mu$eV
Q#39.75
What is the de Broglie wavelength of a red blood cell, with mass that is moving with a speed of Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?
Answer:
The de Broglie wavelength of the blood cell is
$\lambda=\frac{h}{p}=\frac{h}{mv}$
$\lambda = \frac{6.63 \times 10^{-34} \ J.s}{(1.00 \times 10^{-14} \ kg)(4.00 \times 10^{-3} \ m/s)} = 1.66 \times 10^{-17} \ m$
We need not be concerned about wave behavior
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