Particles Behaving as Waves Problems and Solutions 8

Q#39.76

Calculate the energy in electron volts of 

(a) an electron that has de Broglie wavelength 400 nm and 

(b) a photon that has wavelength 400 nm. 

Answer:

An electron and a photon both have the same wavelength. We want to use this fact to calculate the energy of each of them. 

The de Broglie wavelength is $\lambda=\frac{h}{p}$.

The energy of the electron is its kinetic energy, $K = \frac{1}{2}mv^2=\frac{p^2}{2m}$

The energy of the photon is E = hf = $\frac{hc}{\lambda}$

(a) $p = \frac{h}{\lambda}=\frac{6.626 \times 10^{-34} \ J.s}{400 \times 10^{-9} \ m}=1.656 \times 10^{-27} \ kg.m/s$

$E= \frac{p^2}{2m}=\frac{(1.656 \times 10^{-27} \ kg.m/s)^2}{2(9.11 \times 10^{-31} \ kg)}$

$E=1.506 \times 10^{-24} \ J = 9.40 \times 10^{-6} \ eV$

(b) $E =\frac{hc}{\lambda}=\frac{(6.626 \times 10^{-34} \ J.s)(2.998 \times 10^8 \ m/s)}{400 \times 10^{-9} \ m}$ 

$E = 4.966 \times 10^{-19} \ J=3.10 \ eV$

The photon has around 300,000 times as much energy as the electron. 

Q#39.77

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression $\lambda=\frac{h}{p}$ still holds, but we must use the relativistic expression for momentum, $p = mv\sqrt{1-v^2/c^2}$ 

(a) Show that the speed of an electron that has de Broglie wavelength $\lambda$ is

$v=\frac{c}{\sqrt{1+(mc\lambda/h)^2}}$

(b) The quantity $h/mc$ equals $2.426 \times 10^{-12} \ m$ (As we saw in Section 38.3, this same quantity appears in Eq. (38.7), the expression for Compton scattering of photons by electrons.) If $\lambda$ is small compared to $h/mc$, the denominator in the expression found in part (a) is close to unity and the speed is very close to c. In this case it is convenient to write $v=(1 - \Delta)c$ and express the speed of the electron in terms of $\Delta$ rather than v. Find an expression for $\Delta$ valid when $\lambda$ << h/mc. 

[Hint: Use the binomial expansion $(1+z)^n=1+nz+n(n-1)z^2/2+...$, valid for the case |z| << 1.] 

(c) How fast must an electron move for its de Broglie wavelength to be $1.00 \times 10^{-15} \ m$, comparable to the size of a proton? Express your answer in the form $v=(1-\Delta)c$ and state the value of $\Delta$.

Answer:

(a) $\lambda = \frac{h}{p}=\frac{h\left(1-\frac{v^2}{c^2}\right)^{1/2}}{mv}$

square both sides

$\lambda^2 = \frac{h^2\left(1-\frac{v^2}{c^2}\right)}{m^2v^2}$

$\lambda^2 m^2v^2=h^2\left(1-\frac{v^2}{c^2}\right)$

$\lambda^2 m^2v^2=h^2-\left(\frac{h^2v^2}{c^2}\right)$

$(\lambda^2 m^2+\frac{h^2}{c^2})v^2=h^2$

$v^2=\frac{h^2}{\lambda^2 m^2 + \frac{h^2}{c^2}}$

$v^2=\frac{c^2}{\frac{\lambda^2 m^2c^2}{h^2} + 1}$

$v=\sqrt{\frac{c^2}{\left(\frac{\lambda mc}{h}\right)^2 + 1}}$

$v=\frac{c}{\sqrt{\left(\frac{\lambda mc}{h}\right)^2 + 1}}$

(b) $v=\frac{c}{\sqrt{\left(\frac{\lambda mc}{h}\right)^2 + 1}} \approx c\left ( 1-\frac{1}{2}\left ( \frac{mc\lambda}{h} \right )^2 \right )= (1-\Delta)c$

with $\Delta=\frac{1}{2}\left ( \frac{mc\lambda}{h} \right )^2$

(c) $\lambda = 1.00 \times 10^{-15} \ m$ << $\frac{h}{mc}$

$\Delta=\frac{1}{2}\left ( \frac{(9.11 \times 10^{-31})(2.998 \times 10^8 \ m/s)(1.00 \times 10^{-15} \ m)}{(6.626 \times 10^{-34} \ J.s)} \right )^2$

$\Delta = 8.50 \times 10^{-8}$, so

$v = (1 - \Delta)c = (1 - 8.50 \times 10^{-8})c \approx c$

As Δ → 0, v → c and λ → 0.

Q#39.78

Suppose that the uncertainty of position of an electron is equal to the radius of the n = 1 Bohr orbit for hydrogen. Calculate the simultaneous minimum uncertainty of the corresponding momentum component, and compare this with the magnitude of the momentum of the electron in the n = 1 Bohr orbit. Discuss your results.

Answer:

The minimum uncertainty product is $\Delta x \Delta p_x=\frac{\hbar}{2}$. 

$\Delta x = r_1$, where $r_1$ is the radius of the n = 1 Bohr orbit.  In the n = 1 Bohr orbit,

$mv_1r_1=\frac{h}{2\pi}$ and

$p_1= mv_1=\frac{h}{2\pi r_1}$

$\Delta p_x = \frac{\hbar}{2\Delta x}=\frac{\hbar}{2r_1}$

$\Delta p_x=\frac{1.055 \times 10^{-34} \ J.s}{2(0.529 \times 10^{-10} \ m)}=1.0 \times 10^{-24} \ kg.m/s$

This is the same as the magnitude of the momentum of the electron in the n = 1. Bohr orbit. Since the momentum is the same order of magnitude as the uncertainty in the momentum, the uncertainty principle plays a large role in the structure of atoms. 

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