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Chapter 36: Problem 43

Consider an electron whose de Broglie wavelength is equal to the wavelength of green light (about \(550 \mathrm{nm}\) ). a) Treating the electron nonrelativistically, what is its speed? b) Does your calculation confirm that a nonrelativistic treatment is sufficient? c) Calculate the kinetic energy of the electron in \(\mathrm{eV}\).

Short Answer

Expert verified
Answer: The speed of the electron is approximately 1.323 x 10^6 m/s, and its kinetic energy is approximately 4.995 eV.

Step by step solution

01

Identify the de Broglie wavelength formula

To find the speed of the electron, we need to use the de Broglie wavelength formula that relates the wavelength of a particle to its momentum: $$ \lambda = \frac{h}{p} $$ Where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle.
02

Calculate the momentum of the electron

We know the wavelength of the electron is 550 nm, so we can determine its momentum using the de Broglie wavelength formula: $$ p = \frac{h}{\lambda} $$ Using Planck's constant, \(h = 6.626 \times 10^{-34} \mathrm{Js}\), and the wavelength of green light, \(\lambda = 550 \times 10^{-9} \mathrm{m}\): $$ p = \frac{6.626 \times 10^{-34} \mathrm{Js}}{550 \times 10^{-9} \mathrm{m}} = 1.205 \times 10^{-24} \mathrm{kg \cdot m/s} $$
03

Calculate the speed of the electron

The momentum of the electron is related to its mass and speed as follows: $$ p = m_e \cdot v $$ Where \(m_e\) is the mass of the electron and \(v\) is its speed. Rearranging the formula to find the speed, \(v\): $$ v = \frac{p}{m_e} $$ Using the mass of the electron, \(m_e = 9.109 \times 10^{-31} \mathrm{kg}\), and the calculated momentum in step 2: $$ v = \frac{1.205 \times 10^{-24} \mathrm{kg \cdot m/s}}{9.109 \times 10^{-31} \mathrm{kg}} = 1.323 \times 10^6 \mathrm{m/s} $$
04

Check if nonrelativistic treatment is sufficient

A nonrelativistic treatment is sufficient if the electron's speed is much smaller than the speed of light, \(c = 3 \times 10^8 \mathrm{m/s}\). Comparing the electron's speed to the speed of light: $$ \frac{v}{c} \approx \frac{1.323 \times 10^6 \mathrm{m/s}}{3 \times 10^8 \mathrm{m/s}} \approx 0.0044 $$ Since the electron's speed is much smaller than the speed of light, a nonrelativistic treatment is sufficient.
05

Calculate the kinetic energy of the electron

Kinetic energy can be expressed in terms of momentum and mass: $$ K = \frac{p^2}{2m_e} $$ Using the calculated momentum in step 2 and the mass of the electron: $$ K = \frac{(1.205 \times 10^{-24} \mathrm{kg \cdot m/s})^2}{2 \times 9.109 \times 10^{-31} \mathrm{kg}} = 8.002 \times 10^{-19} \mathrm{J} $$ To convert the kinetic energy to electronvolts (eV), we divide the energy in joules by the elementary charge, \(e = 1.602 \times 10^{-19} \mathrm{C}\): $$ K_{\mathrm{eV}} = \frac{8.002 \times 10^{-19} \mathrm{J}}{1.602 \times 10^{-19} \mathrm{C}} = 4.995 \mathrm{eV} $$ Hence, the kinetic energy of the electron is approximately 4.995 eV.

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Most popular questions from this chapter

A photovoltaic device uses monochromatic light of wavelength 700 . \(\mathrm{nm}\) that is incident normally on a surface of area \(10.0 \mathrm{~cm}^{2}\). Calculate the photon flux rate if the light intensity is \(0.300 \mathrm{~W} / \mathrm{cm}^{2}\).

In baseball, a \(100 .-\mathrm{g}\) ball can travel as fast as \(100 . \mathrm{mph}\). What is the de Broglie wavelength of this ball? The Voyager spacecraft, with a mass of about \(250 . \mathrm{kg}\), is currently travelling at \(125,000 \mathrm{~km} / \mathrm{h}\). What is its de Broglie wavelength?

X-rays of wavelength \(\lambda=0.120 \mathrm{nm}\) are scattered from carbon. What is the Compton wavelength shift for photons detected at \(90.0^{\circ}\) angle relative to the incident beam?

Now consider de Broglie waves for a (relativistic) particle of mass \(m\), momentum \(p=m v \gamma\), and total energy \(E=m c^{2} \gamma\), with \(\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}\). The waves have wavelength \(\lambda=h / p\) and frequency \(f=E / h\) as before, but with the relativistic momentum and energy. a) Calculate the dispersion relation for these waves. b) Calculate the phase and group velocities of these waves. Now which corresponds to the classical velocity of the particle?

An accelerator boosts a proton's kinetic energy so that the de Broglie wavelength of the proton is \(3.5 \cdot 10^{-15} \mathrm{~m}\) What are the momentum and energy of the proton?
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