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(a) What is the de Broglie wavelength of an electron accelerated from rest through a potential increase of 125 \(\mathrm{V} ?\) (b) What is the de Broglie wavelength of an alpha particle \((q=+2 e,\) \(m=6.64 \times 10^{-27} \mathrm{kg} )\) accelerated from rest through a potential drop of 125 \(\mathrm{V} ?\)

Short Answer

Expert verified
(a) The de Broglie wavelength of the electron is \(1.24 \times 10^{-10} \ \text{m}\). (b) The de Broglie wavelength of the alpha particle is \(9.1 \times 10^{-11} \ \text{m}\).

Step by step solution

01

Determine the electron's kinetic energy

When an electron is accelerated through a potential difference, its kinetic energy gained is equal to the work done on it. This can be calculated as \( KE = eV \), where \( e \) is the elementary charge \( (1.6 \times 10^{-19} \ \text{C}) \) and \( V \) is the voltage \( (125 \ \text{V}) \). Thus, \( KE = 1.6 \times 10^{-19} \times 125 = 2.0 \times 10^{-17} \ \text{J} \).
02

Calculate the momentum of the electron

The kinetic energy \( KE \) is also given by \( \frac{1}{2}mv^2 = \frac{p^2}{2m} \), where \( p \) is the momentum. Solving for \( p \), we have \( p = \sqrt{2m \cdot KE} \). The mass of an electron \( m = 9.11 \times 10^{-31} \ \text{kg} \). Thus, \( p = \sqrt{2 \times 9.11 \times 10^{-31} \times 2.0 \times 10^{-17}} = 5.34 \times 10^{-24} \ \text{kg} \cdot \text{m/s} \).
03

Find the de Broglie wavelength of the electron

The de Broglie wavelength \( \lambda \) is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \ \text{J} \cdot \text{s} \). Substituting the values, \( \lambda = \frac{6.626 \times 10^{-34}}{5.34 \times 10^{-24}} = 1.24 \times 10^{-10} \ \text{m} \).
04

Determine the alpha particle's kinetic energy

The kinetic energy of the alpha particle is \( qV \), where \( q = 2e = 3.2 \times 10^{-19} \ \text{C} \) and \( V = 125 \ \text{V} \). Therefore, \( KE = 3.2 \times 10^{-19} \times 125 = 4.0 \times 10^{-17} \ \text{J} \).
05

Calculate the momentum of the alpha particle

Similar to the electron, \( p_{\text{alpha}} = \sqrt{2m_{\text{alpha}} \cdot KE} \). With \( m_{\text{alpha}} = 6.64 \times 10^{-27} \ \text{kg} \), \( p_{\text{alpha}} = \sqrt{2 \times 6.64 \times 10^{-27} \times 4.0 \times 10^{-17}} = 7.28 \times 10^{-24} \ \text{kg} \cdot \text{m/s} \).
06

Find the de Broglie wavelength of the alpha particle

For the alpha particle, \( \lambda_{\text{alpha}} = \frac{h}{p_{\text{alpha}}} \). Substituting the values, we get \( \lambda_{\text{alpha}} = \frac{6.626 \times 10^{-34}}{7.28 \times 10^{-24}} = 9.1 \times 10^{-11} \ \text{m} \).

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Key Concepts

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

Understanding Electron Kinetic Energy
When an electron is accelerated from rest through a potential difference of 125 volts, it gains kinetic energy. This is a result of the work done on the electron by the electric field generated by the potential difference. The formula for the kinetic energy (KE) acquired by the electron is given by \( KE = eV \), where \( e \) is the elementary charge \( 1.6 \times 10^{-19} \ \text{C} \). With \( V \) being the voltage (125 V), the kinetic energy becomes \( KE = 1.6 \times 10^{-19} \times 125 = 2.0 \times 10^{-17} \ \text{J} \).\\The equation essentially states that the electric potential energy lost is converted into kinetic energy of the electron. This is a crucial concept often tested in physics, helping us understand how particles gain energy in electric fields. It shows how electric fields can be used to accelerate charged particles, a principle that is fundamental in designing devices like particle accelerators.\\Understanding electron kinetic energy can open doors to learning about various phenomena, from the working of cathode rays to the concept of voltage in circuits.
Exploring Alpha Particle Kinetic Energy
In a similar manner, when an alpha particle, with a charge of \( q = +2e \), is subject to the same voltage, it gains kinetic energy due to the electric field's work on it. The calculation for its kinetic energy, however, considers its charge as \( q = 2 \times e = 3.2 \times 10^{-19} \ \text{C} \).\\The alpha particle's kinetic energy as it accelerates through the 125 V potential drop is then \( KE = qV = 3.2 \times 10^{-19} \times 125 = 4.0 \times 10^{-17} \ \text{J} \).\\An alpha particle, consisting of two protons and two neutrons, is far heavier than an electron. Hence, though it carries more energy for the same potential difference, its behavior in electric fields is remarkably different. This differentiates the dynamics of alpha particles from those of lighter electrons and plays a significant role in nuclear physics and other advanced domains.\\These insights into alpha particle kinetic energy help us further appreciate how different entities interact within electromagnetic fields, shaping much of the modern scientific understanding.
Breaking Down Momentum Calculation
Momentum is a crucial physical quantity that provides insight into an object's motion, and for particles like electrons and alpha particles, it can be linked directly to kinetic energy. In this context, the momentum \( p \) of a particle can be calculated using the relation \( KE = \frac{p^2}{2m} \), which can be manipulated to \( p = \sqrt{2m \cdot KE} \).\\For the electron, with mass \( m = 9.11 \times 10^{-31} \ \text{kg} \) and kinetic energy \( KE = 2.0 \times 10^{-17} \ \text{J} \), the momentum comes out as \( p = \sqrt{2 \times 9.11 \times 10^{-31} \times 2.0 \times 10^{-17}} = 5.34 \times 10^{-24} \ \text{kg} \cdot \text{m/s} \).\\Similarly, for the alpha particle with mass \( m_{\text{alpha}} = 6.64 \times 10^{-27} \ \text{kg} \) and kinetic energy \( KE = 4.0 \times 10^{-17} \ \text{J} \), its momentum calculated is \( p_{\text{alpha}} = \sqrt{2 \times 6.64 \times 10^{-27} \times 4.0 \times 10^{-17}} = 7.28 \times 10^{-24} \ \text{kg} \cdot \text{m/s} \).\\Understanding momentum is essential as it is a conserved quantity in closed systems and plays a critical role when predicting and analyzing particle interactions in physics projects and experiments. A thorough grasp of momentum and its calculation strengthens your ability to understand more complex scenarios encountered in advanced physics.

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Most popular questions from this chapter

Particle \(A\) is described by the wave function \(\psi(x, y, z)\) . Particle \(B\) is described by the wave function \(\psi(x, y, z) e^{i \phi}\) , where \(\phi\) is a real constant. How does the probability of finding particle \(A\) within a volume \(d V\) around a certain point in space compare with the probability of finding particle \(B\) within this same volume?

A particle is described by the normalized wave function \(\psi(x, y, z)=A x e^{-\alpha x^{2}} e^{-\beta \beta} e^{-\gamma x^{2}},\) where \(A, \alpha, \beta,\) and \(\gamma\) are all real, positive constants. The probability that the particle will be found in the infinitesimal volume \(d x d y d z\) centered at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) is \(\left|\psi\left(x_{0}, y_{0}, z_{0}\right)\right|^{2} d x d y d z\) (a) At what value of \(x_{0}\) is the particle most likely to be found? (b) Are there values of \(x_{0}\) for which the probability of the particle being found is zero? If so, at what \(x_{0} ?\)

The wave nature of particles results in the quantum- mechanical situation that a particle confined in a box can assume only wavelengths that result in standing waves in the box, with nodes at the box walls, (a) Show that an electron confined in a one- dimensional box of length \(L\) will have energy levels given by $$E_{n}=\frac{n^{2} h^{2}}{8 m L^{2}}$$ (Hint: Recall that the relationship between the de Broglic wave- length and the speed of a nonrelativistic particle is \(m v=h / \lambda\) . The energy of the particle is \(\frac{1}{2} m v^{2}, )(b)\) If a hydrogen atom is modeled as a one-dimensional box with length equal to the Bohr radius, what is the energy (in electron volts) of the lowest energy level of the electron?

By extremely careful measurement, you determine the \(x\) -coondinate of a car's center of mass with an uncertainty of only 1.00\(\mu \mathrm{m}\) . The car has a mass of 1200 \(\mathrm{kg}\) . (a) What is the minimum uncertainty in the \(x\) -component of the velocity of the car's center of Mass as presuribed by the Heisenberg uncertainty principle? (b) Does the uncertainty principle impose a practical limit on our ability to make simultaneous measurements of the positions and velocities of ordinary objects like cars, books, and people? Explain.

Imagine another universe in which the value of Planck's constant is \(0.0663 \mathrm{J} \cdot \mathrm{s},\) but in which the physical laws and all other physical constants are the same as in our universe. In this universe, two physics students are playing catch. They are 12 \(\mathrm{m}\) apart, and one throws a \(0.25-\mathrm{kg}\) ball directly toward the other with a speed of 6.0 \(\mathrm{m} / \mathrm{s}\) (a) What is the uncertainty in the ball's horizontal momentum, in a direction perpendicular to that in which it is being thrown, if the student throwing the ball knows that it is located within a cube with volume 125 \(\mathrm{cm}^{3}\) at the time she throws it? (b) By what horizontal distance could the ball miss the second student?

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