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Chapter 42: Problem 41

Compute the de Broglie wavelength of an electron that has been accelerated through a potential difference of \(9.0 \mathrm{kV}\). Ignore relativistic effects.

Short Answer

Expert verified
The de Broglie wavelength of the electron is approximately \(1.3 \times 10^{-10}\) meters.

Step by step solution

01

Calculate the Kinetic Energy

The kinetic energy gained by the electron when it is accelerated through a potential difference of 9.0 kV can be calculated using the formula: \[ KE = e \cdot V \]where \( e \approx 1.6 \times 10^{-19} \text{C} \) is the charge of the electron and \( V = 9.0 \text{kV} = 9.0 \times 10^3 \text{V} \). Thus, the kinetic energy is:\[ KE = 1.6 \times 10^{-19} \times 9.0 \times 10^3 = 1.44 \times 10^{-15} \text{J} \]
02

Determine the Momentum of the Electron

The momentum \( p \) of the electron can be related to its kinetic energy by the expression:\[ KE = \frac{p^2}{2m} \]Solving for \( p \), we get:\[ p = \sqrt{2m \cdot KE} \]where \( m = 9.11 \times 10^{-31} \text{kg} \) is the mass of the electron. Substituting the known values:\[ p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.44 \times 10^{-15}} \approx 5.1 \times 10^{-24} \text{kg m/s} \]
03

Apply the de Broglie Wavelength Formula

Using the de Broglie wavelength formula \( \lambda = \frac{h}{p} \) where \( h = 6.63 \times 10^{-34} \text{J s} \) is Planck's constant, we can find the wavelength:\[ \lambda = \frac{6.63 \times 10^{-34}}{5.1 \times 10^{-24}} \approx 1.3 \times 10^{-10} \text{m} \]
04

Conclusion

Rounding it, the de Broglie wavelength of the electron, ignoring relativistic effects, is approximately \(1.3 \times 10^{-10} \text{m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Acceleration
When an electron is accelerated, it is moved faster due to an external force, such as an electric field. This results in electrons gaining energy as they move through a potential difference. In physics, a potential difference is often provided in volts (V). If an electron is accelerated through a potential difference of 9.0 kV, its velocity increases. This is because the electric field does work on the electron, converting electric potential energy into kinetic energy. The higher the potential difference, the more kinetic energy the electron gains, making electron acceleration crucial in understanding how the energy changes are involved.
Kinetic Energy
Kinetic energy refers to the energy that an object possesses due to its motion. For an electron accelerated through an electric potential difference, this energy can be calculated using the formula: \[ KE = e \times V \]where \( e \approx 1.6 \times 10^{-19} \text{C} \) is the charge of an electron, and \( V \) is the potential difference. For an electron accelerated through 9.0 kV, the kinetic energy can be found by substituting these values into the formula. Kinetic energy is a critical factor because it helps us determine how fast the electron is moving, which is important in calculating momentum and further understanding the electron's behavior.
Momentum
Momentum is the measure of the motion of an object, which depends on both its mass and velocity. For electrons, momentum \( p \) can be related to kinetic energy using:\[ KE = \frac{p^2}{2m} \]Here, \( m \) represents the mass of the electron. By rearranging the formula, we can solve for momentum:\[ p = \sqrt{2m \cdot KE} \]This formula shows that the greater the kinetic energy, the larger the momentum. Electron momentum is significant in fields such as quantum mechanics, as it helps determine electron behavior, including its de Broglie wavelength.
Planck's Constant
Planck's constant \( h \) is a fundamental constant in physics, representing the proportionality between the energy of a photon and its frequency. It is an essential element in quantum physics, with a value of approximately \( 6.63 \times 10^{-34} \text{J s} \). When considering an electron's wave-like properties, Planck's constant is used in the de Broglie wavelength formula:\[ \lambda = \frac{h}{p} \]where \( \lambda \) is the wavelength and \( p \) is the momentum of the electron. Planck鈥檚 constant helps link particle and wave characteristics of particles, crucial in quantum physics and understanding electron behavior on a microscopic scale.

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Most popular questions from this chapter

Suppose that a 3.64-nm photon moving in the \(+x\) -direction collides head-on with a \(2 \times 10^{5} \mathrm{~m} / \mathrm{s}\) electron moving in the \(-x\) direction. If the collision is perfectly elastic, find the conditions after collision. From the law of conservation of momentum, Momentum before = Momentum after $$ \frac{h}{\lambda_{0}}-m v_{0}=\frac{h}{\lambda}-m v $$ But, from \(\underline{\text { Problem } 42.9,} h / \lambda_{0}=m u\) in this case. Hence, \(h / \lambda=m v\). Also, for a perfectly elastic collision, $$ \begin{array}{l} \text { KE before }=\text { KE after } \\ \frac{h c}{\lambda_{0}}+\frac{1}{2} m v_{0}^{2}=\frac{h c}{\lambda}+\frac{1}{2} m v^{2} \end{array} $$ Using the facts that \(h / \lambda_{0}=m v_{0}\) and \(h / \lambda=m v\), we find $$ v_{0}\left(\mathrm{c}+\frac{1}{2} v_{0}\right)=v\left(\mathrm{c}+\frac{1}{2} v\right) $$ Therefore, \(v=\mathrm{u}_{0}\) and the electron moves in the \(+\chi\) -direction with its original speed. Because \(h / \lambda=m v=m u_{0}\), the photon also "rebounds," and with its original wavelength.

What is the wavelength of light in which the photons have an energy of \(600 \mathrm{eV}\) ?

What potential difference must be applied to stop the fastest photoelectrons emitted by a nickel surface under the action of ultraviolet light of wavelength \(200 \mathrm{~nm}\) ? The work function of nickel is \(5.01 \mathrm{eV}\). $$ \mathrm{E}=\frac{h \mathrm{c}}{\lambda}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{2000 \times 10^{-10} \mathrm{~m}}=9.95 \times 10^{-19} \mathrm{~J}=6.21 \mathrm{eV} $$ Then, from the photoelectric equation, the energy of the fastest emitted electron is $$ 6.21 \mathrm{eV}-5.01 \mathrm{eV}=1.20 \mathrm{eV} $$ Hence, a negative retarding potential of \(1.20 \mathrm{~V}\) is required. This is the stopping potential.

What is the work function of sodium metal if the photoelectric threshold wavelength is \(680 \mathrm{~nm}\) ?

To break a chemical bond in the molecules of human skin and thus cause sunburn, a photon energy of about \(3.50 \mathrm{eV}\) is required. To what wavelength does this correspond? $$ \lambda=\frac{h \mathrm{c}}{\mathrm{E}}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{(3.50 \mathrm{eV})\left(1.602 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}=354 \mathrm{~nm} $$ Ultraviolet radiation causes sunburn.
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