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High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda = h/p\) 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 $$v = \frac{c}{\sqrt1+(mc\lambda/h)^2} $$ (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 \(v\) 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 rather than v. Find an expression for \(\delta\) valid when \(\lambda \ll h mc\). [\(Hint:\) Use the binomial expansion (1 + \(z)^n = 1 + nz + [n(n - 1)z^2/2] + \cdots\) g, valid for the case 0 z 0 6 1.4 (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 \(\Delta\)

Short Answer

Expert verified
(a) For the speed, use \( v = \frac{c}{\sqrt{1 + (mc\lambda/h)^2}} \). (b) \( \Delta \approx 8.5 \times 10^{-8} \). (c) The electron's speed is \( v \approx (1 - 8.5 \times 10^{-8})c \).

Step by step solution

01

Relate de Broglie Wavelength to Relativistic Momentum

The de Broglie wavelength is given by\( \lambda = \frac{h}{p} \). Using the relativistic expression for momentum \( p = \frac{mv}{\sqrt{1 - v^2/c^2}} \), plug this into the de Broglie equation to get\[ \lambda = \frac{h}{mv}\sqrt{1 - \frac{v^2}{c^2}} \].
02

Solve for Velocity v in Terms of Wavelength

Rearranging the equation from the previous step gives:\[ v = \frac{h}{m\lambda}\sqrt{1 - \frac{v^2}{c^2}} \].Isolating \( v \), we obtain:\[ v = \frac{c}{\sqrt{1 + \left(\frac{mc\lambda}{h}\right)^2}} \],as required.
03

Evaluate the Limit When Wavelength is Much Smaller than h/mc

Since \( \lambda \ll h/mc \), then \( \left(\frac{mc\lambda}{h}\right)^2 \ll 1 \). Use the binomial expansion\[ \frac{1}{\sqrt{1 + x}} \approx 1 - \frac{x}{2} \]for small \( x \).This implies:\[ \frac{c}{\sqrt{1 + \left(\frac{mc\lambda}{h}\right)^2}} \approx c\left(1 - \frac{1}{2}\left(\frac{mc\lambda}{h}\right)^2\right) \].Thus, \( v = (1 - \Delta)c \) and \( \Delta = \frac{1}{2}\left(\frac{mc\lambda}{h}\right)^2 \).
04

Calculate Electron Speed for Given Wavelength

Given the wavelength \( \lambda = 1.00 \times 10^{-15} \text{ m} \), compute \( \Delta \) using previously derived expression\[ \Delta = \frac{1}{2}\left(\frac{mc\lambda}{h}\right)^2 \].Substitute \( \frac{h}{mc} = 2.426 \times 10^{-12} \text{ m} \), then\[ \Delta = \frac{1}{2}\left(\frac{1.00 \times 10^{-15}}{2.426 \times 10^{-12}}\right)^2 \approx \frac{1}{2}\left(4.12 \times 10^{-4}\right)^2 = \frac{1}{2} \times (1.7 \times 10^{-7}) \approx 8.5 \times 10^{-8} \].

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Key Concepts

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

de Broglie Wavelength
The de Broglie wavelength is an essential concept in quantum mechanics. It bridges the wave-like and particle-like properties of matter. This wavelength is defined specifically for any moving particle, such as an electron, through the relation: \( \lambda = \frac{h}{p} \), where \( \lambda \) is the de Broglie wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle. This formula tells us that every moving object has a wave associated with it. The smaller the object or the higher its speed, the more significant its wave nature becomes, creating implications for particles like electrons.
When electrons move at very high speeds, especially close to the speed of light, we must take relativistic effects into account while calculating their de Broglie wavelength. The usual momentum \( p = mv \) gets replaced by the relativistic momentum formula, emphasizing the need for understanding both the wave nature and relativistic aspects of fast-moving particles.
Relativistic Momentum
As particles move at speeds comparable to the speed of light, Einstein's theory of relativity alters how we understand momentum. Instead of using classical mechanics, where momentum is simply the product of mass and velocity, relativistic momentum must be considered:
  • The formula for relativistic momentum is \( p = \frac{mv}{\sqrt{1 - v^2/c^2}} \).
  • Here, \( m \) is the mass of the particle, \( v \) is its velocity, and \( c \) is the speed of light.
As objects approach the speed of light, the denominator \( \sqrt{1 - v^2/c^2} \) becomes much smaller, significantly increasing the value of \( p \).
This adjustment in momentum is crucial for calculating properties like the de Broglie wavelength for high-speed electrons.
The influence of relativistic momentum becomes critical because it ensures that speeds above the speed of light are mathematically restricted, aligning with the principles of Einstein's relativity. Thus, understanding this momentum is pivotal when examining particles in high-energy physics.
Binomial Expansion
The binomial expansion is a mathematical method useful for approximating expressions involving terms raised to a power. It particularly helps when dealing with small quantities compared to unity, allowing the simplification of complex formulas. The binomial theorem states that:
  • The expression \((1 + z)^n\) can be expanded to \(1 + nz + \frac{n(n-1)}{2}z^2 + \cdots\).
In our context, when calculating relativistic speeds for electrons with a de Broglie wavelength much smaller than \(h/mc\), the binomial expansion allows us to approximate terms like \(\sqrt{1 + x}\).
This approximation \(\frac{1}{\sqrt{1 + x}} \approx 1 - \frac{x}{2}\) when \( x \ll 1 \), simplifies calculations by reducing the complexity of terms. For electron speeds near the speed of light, this simplification helps derive expressions for speed \(v\) as \((1 - \Delta)c\).
In cases like these, simplicity and accuracy are key, making the binomial expansion a valuable tool in theoretical physics for reducing formidable equations to manageable calculations.

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Most popular questions from this chapter

A triply ionized beryllium ion, Be\(^{3+}\) (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (a) What is the ground-level energy of Be\(^{3+}\)? How does this compare to the ground-level energy of the hydrogen atom? (b) What is the ionization energy of Be\(^{3+}\)? How does this compare to the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the \(n\) = 2 to \(n\) = 1 transition is 122 nm (see Example 39.6). What is the wavelength of the photon emitted when a Be\(^{3+}\) ion undergoes this transition? (d) For a given value of \(n\), how does the radius of an orbit in Be\(^{3+}\) compare to that for hydrogen?

A scientist has devised a new method of isolating individual particles. He claims that this method enables him to detect simultaneously the position of a particle along an axis with a standard deviation of 0.12 nm and its momentum component along this axis with a standard deviation of 3.0 \(\times\) 10\(^{-25}\) kg \(\bullet\) m/s. Use the Heisenberg uncertainty principle to evaluate the validity of this claim.

In the second type of helium-ion microscope, a 1.2-MeV ion passing through a cell loses 0.2 MeV per \(\mu\)m of cell thickness. If the energy of the ion can be measured to 6 keV, what is the smallest difference in thickness that can be discerned? (a) 0.03 \(\mu\)m; (b) 0.06 \(\mu\)m; (c) 3 \(\mu\)m; (d) 6 \(\mu\)m.

A certain atom has an energy state 3.50 eV above the ground state. When excited to this state, the atom remains for 2.0 \(\mu\)s, on average, before it emits a photon and returns to the ground state. (a) What are the energy and wavelength of the photon? (b) What is the smallest possible uncertainty in energy of the photon?

An atom with mass \(m\) emits a photon of wavelength \(\lambda\). (a) What is the recoil speed of the atom? (b) What is the kinetic energy \(K\) of the recoiling atom? (c) Find the ratio \(K/E\), where \(E\) is the energy of the emitted photon. If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths? (d) Calculate \(K\) (in electron volts) and \(K/E\) for a hydrogen atom (mass 1.67 \(\times\) 10\(^{-27}\)kg) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process?

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