Q#39.67
(a) Write the Planck distribution law in terms of the frequency f, rather than the wavelength $\lambda$ to obtain I(f).
(b) Show that
$\int_{0}^{\infty }I{(\lambda)}\ d\lambda=\frac{2\pi^5k^4}{15c^2h^3}T^4$
where I($\lambda$) is the Planck distribution formula of Eq. (39.24). (Hint: Change the integration variable from $\lambda$ to f. You will need to use the following tabulated integral:
$\int_{0}^{\infty }\frac{x^3}{e^{\alpha x}-1}\ dx=\frac{1}{24}\left(\frac{2\pi}{\alpha}\right)^4$
(c) The result of part (b) is I and has the form of the Stefan– Boltzmann law, $I = \sigma T^4$ (Eq. 39.19). Evaluate the constants in part (b) to show that has the value given in Section 39.5.
Answer:
(a) $I(\lambda )=\frac{2\pi hc^2}{\lambda^ 5\left ( e^{hc/\lambda kT}-1 \right )}$, but $\lambda = \frac{c}{f}$,
so, the Planck distribution law in terms of the frequency f is
$I(f )=\frac{2\pi hc^2}{(c/f)^ 5\left ( e^{hc/(c/f) kT}-1 \right )}$
$I(f )=\frac{2\pi hc^2f^5}{c^ 5\left ( e^{hf/kT}-1 \right )}$
$I(f )=\frac{2\pi hf^5}{c^3\left ( e^{hf/kT}-1 \right )}$
(b) $\int_{0}^{\infty }I{(\lambda)}\ d\lambda=\int_{\infty }^{0}I(f) \ df\left ( \frac{-c}{f^2} \right )$
$\int_{0}^{\infty }I{(\lambda)}\ d\lambda=\int_{\infty }^{0}\frac{2\pi hf^5}{c^3\left ( e^{hf/kT}-1 \right )} \ df\left ( \frac{-c}{f^2} \right )$
$\int_{\infty}^{0 }I{(\lambda)}\ d\lambda=\int_{\infty }^{0}\frac{2\pi hf^3}{c^2\left ( e^{hf/kT}-1 \right )} \ df $
suppose, $\frac{hf}{kT}=x$, so
$\int_{\infty}^{0 }I{(\lambda)}\ d\lambda=\int_{\infty }^{0}\frac{2\pi hf^3 \times \frac{h^2}{h^2}\times \left (\frac{kT}{kT} \right )^3}{c^2\left ( e^{hf/kT}-1 \right )} \ df$
$\int_{\infty}^{0 }I{(\lambda)}\ d\lambda=\int_{\infty }^{0}\frac{2\pi \frac{k^3T^3}{h^2} \left (\frac{hf}{kT} \right )^3}{c^2\left ( e^{hf/kT}-1 \right )} \ df $
because, $\frac{hf}{kT}=x$, so $\frac{h}{kT} df=dx$ → $df=\frac{kT}{h}dx$
$\int_{\infty}^{0 }I{(\lambda)}\ d\lambda=\int_{\infty }^{0}\frac{2\pi \frac{k^3T^3}{h^2} \left (\frac{hf}{kT} \right )^3}{c^2\left ( e^{hf/kT}-1 \right )} \ \left ( \frac{kT}{h} \right )dx $
$\int_{\infty}^{0 }I{(\lambda)}\ d\lambda=\frac{2\pi (kT)^4}{h^3c^2}\int_{\infty }^{0}\frac{x^3}{e^x-1}dx$
$\int_{\infty}^{0 }I{(\lambda)}\ d\lambda=\frac{2\pi (kT)^4}{h^3c^2}\frac{1}{240}(2\pi)^4$
$\int_{\infty}^{0 }I{(\lambda)}\ d\lambda=\frac{(2\pi)^5 (kT)^4}{2^4\times 15h^3c^2}$
$\int_{\infty}^{0 }I{(\lambda)}\ d\lambda=\frac{2\pi^5 (kT)^4}{15h^3c^2}$
(c) The expression $\frac{2\pi^5 k^4}{15h^3c^2}=\sigma$ as shown in Eq. (39.28). Plugging in the values for the constants we get
$\sigma=5.67 \times 10^{-8} \ W.m^{-2}.K^{-4}$
Q#39.68
A beam of 40-eV electrons traveling in the +x-direction passes through a slit that is parallel to the y-axis and 5.0 $\mu$m wide. The diffraction pattern is recorded on a screen 2.5 m from the slit.
(a) What is the de Broglie wavelength of the electrons?
(b) How much time does it take the electrons to travel from the slit to the screen?
(c) Use the width of the central diffraction pattern to calculate the uncertainty in the y-component of momentum of an electron just after it has passed through the slit.
(d) Use the result of part (c) and the Heisenberg uncertainty principle (Eq. 39.29 for y) to estimate the minimum uncertainty in the y-coordinate of an electron just after it has passed through the slit. Compare your result to the width of the slit.
Answer:
$\lambda = \frac{h}{p}=\frac{h}{\sqrt{2mE}}$. From Chapter 36, if λ << a then the width w of the central maximum is
$w=\frac{2R\lambda}{a}$
where R = 2.5 m and a is the width of the slit.
$v_x=\sqrt{\frac{2E}{m}}$, since the beam is traveling in the x-direction and $\Delta v_y$y << $v_x$
(a) $\lambda = \frac{h}{\sqrt{2mK}}$
$\lambda = \frac{6.626 \times 10^{-34} \ J.s}{\sqrt{2(9.11 \times 10^{-31} \ kg)(40 \ eV)(1.60 \times 10^{-19} \ J/eV)}}$
$\lambda = 1.94 \times 10^{-10} \ m$
(b) $t = \frac{R}{v}=\frac{R}{\sqrt{2E/m}}$
$t=\frac{2.5 \ m}{\sqrt{\frac{(2)(40 \ eV)(1.60 \times 10^{-19} \ J/eV)}{9.11 \times 10^{-31} \ kg}}}$
$t=6.67 \times 10^{-7} \ s$
(c) The width w is $w=2R\frac{\lambda}{a}$ and $w=\Delta v_y t=\frac{\Delta p_y t}{m}$
where t is the time found in part (b) and a is the slit width. Combining the expressions for w,
$\Delta p_y=\frac{2m\lambda R}{at}$
$\Delta p_y=\frac{2(9.11 \times 10^{-31} \ kg)(1.94 \times 10^{-10} \ m)(2.5 \ m)}{(5.0 \times 10^{-6} \ m)(6.67 \times 10^{-7} \ s)}$
$\Delta p_y = 2.65 \times 10^{-28} \ kg.m/s$
(d) $\Delta y = \frac{h}{4\pi \Delta p_y}$
$\Delta y = \frac{6.63 \times 10^{-34} \ J.s}{4\pi (2.65 \times 10^{-28} \ kg.m/s)}$
$\Delta y = 1.89 \times 10^{-7} \ m \approx 0.20 \ \mu$m, which is the same order of magnitude of the width of the slit.
For these electrons $\lambda = 1.94 \times 10^{-10} \ m$.
This is much smaller than a and the approximate expression $w = \frac{2R\lambda}{a}$ is very accurate. Also,
$v_x=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(40 \ eV)(1.60 \times 10^{-19} \ J/eV)}{9.11 \times 10^{-31} \ kg}}$
$v_x=3.75 \times 10^6 \ m/s$
$\Delta v_y=\frac{\Delta p_y}{m}=\frac{2.65 \times 10^{-28} \ kg.m/s}{9.11 \times 10^{-31} \ kg} = 2.9 \times 10^2 \ m/s$, so it is the case that $v_x>>\Delta v_y$
Q#39.69
(a) What is the energy of a photon that has wavelength 0.10 $\mu$m?
(b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10 $\mu$m in diameter? What is the speed of these electrons?
(c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?
Answer:
For a photon $E= \frac{hc}{\lambda}$
For a particle with mass, $p=\frac{h}{\lambda}$ and $E = \frac{p^2}{2m}=q \Delta V$,
where ΔV is the accelerating voltage. To exhibit wave nature when passing through an opening, the de Broglie wavelength of the particle must be comparable with the width of the opening.
(a) the energy of a photon that has wavelength 0.10 $\mu$m is
$E= \frac{hc}{\lambda}=\frac{(6.626 \times 10^{-34} \ J.s)(3.00 \times 10^8 \ m/s)}{0.10 \times 10^{-6} \ m}$
$E= 1.99 \times 10^{-18} J = 12.4 \ eV$
(b) Find E for an electron with $\lambda = 0.10 \times 10^{-6} \ m$,
$p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34} \ J.s}{0.10 \times 10^{-6} \ m}=6.626 \times 10^{-27} \ kg.m/s$
$E = \frac{p^2}{2m}= \frac{(6.626 \times 10^{-27} \ kg.m/s)^2}{2(9.11 x 10^{-31} \ kg)}$
$E = 2.41 \times 10^{-23} \ J = 1.5 \times 10^{-4} \ eV$
$E = q\Delta V$
$1.5 \times 10^{-4} \ eV=e \Delta V$
$\Delta V = 1.5 \times 10^{-4} \ V$
$v = \frac{p}{m}=\frac{6.626 \times 10^{-27} \ kg.m/s}{9.11 \times 10^{-31} \ kg} = 7.3 \times 10^3 \ m/s$
(c) Same λ so same p.
$E = \frac{p^2}{2m}= \frac{(6.626 \times 10^{-27} \ kg.m/s)^2}{2(1.673 x 10^{-27} \ kg)}$
$E = 8.2 \times 10^{-8} \ eV$ and $\Delta V = 8.2 \times 10^{-8} \ V$
$v = \frac{p}{m}=\frac{6.626 \times 10^{-27} \ kg.m/s}{1.673 \times 10^{-27} \ kg} = 4.0 \ m/s$
A proton must be traveling much slower than an electron in order to have the same de Broglie wavelength.
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