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High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression λ=h/p still holds, but we must use the relativistic expression for momentum, p=mv/1v2/c2. (a) Show that the speed of an electron that has de Broglie wavelength λ is v=c1+(mcλ/h)2 (b) The quantity h/mc equals 2.426 × 1012 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 λ 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=(1Δ)c and express the speed of the electron in terms of rather than v. Find an expression for δ valid when λhmc. [Hint: Use the binomial expansion (1 + z)n=1+nz+[n(n1)z2/2]+ 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 × 1015 m, comparable to the size of a proton? Express your answer in the form v=(1Δ)c, and state the value of Δ

Short Answer

Expert verified
(a) For the speed, use v=c1+(mcλ/h)2. (b) Δ8.5×108. (c) The electron's speed is v(18.5×108)c.

Step by step solution

01

Relate de Broglie Wavelength to Relativistic Momentum

The de Broglie wavelength is given byλ=hp. Using the relativistic expression for momentum p=mv1v2/c2, plug this into the de Broglie equation to getλ=hmv1v2c2.
02

Solve for Velocity v in Terms of Wavelength

Rearranging the equation from the previous step gives:v=hmλ1v2c2.Isolating v, we obtain:v=c1+(mcλh)2,as required.
03

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

Since λh/mc, then (mcλh)21. Use the binomial expansion11+x1x2for small x.This implies:c1+(mcλh)2c(112(mcλh)2).Thus, v=(1Δ)c and Δ=12(mcλh)2.
04

Calculate Electron Speed for Given Wavelength

Given the wavelength λ=1.00×1015 m, compute Δ using previously derived expressionΔ=12(mcλh)2.Substitute hmc=2.426×1012 m, thenΔ=12(1.00×10152.426×1012)212(4.12×104)2=12×(1.7×107)8.5×108.

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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: λ=hp, where λ 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=mv1v2/c2.
  • 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 1v2/c2 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+n(n1)2z2+.
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 1+x.
This approximation 11+x1x2 when x1, 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 (1Δ)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

The radii of atomic nuclei are of the order of 5.0 × 1015 m. (a) Estimate the minimum uncertainty in the momentum of an electron if it is confined within a nucleus. (b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the kinetic energy of an electron confined within a nucleus. (c) Compare the energy calculated in part (b) to the magnitude of the Coulomb potential energy of a proton and an electron separated by 5.0 × 1015 m. On the basis of your result, could there be electrons within the nucleus? (Note: It is interesting to compare this result to that of Problem 39.72.)

The neutral pion (π0) is an unstable particle produced in high-energy particle collisions. Its mass is about 264 times that of the electron, and it exists for an average lifetime of 8.4 × 1017 s before decaying into two gamma-ray photons. Using the relationship E=mc2 between rest mass and energy, find the uncertainty in the mass of the particle and express it as a fraction of the mass.

(a) A particle with mass m has kinetic energy equal to three times its rest energy. What is the de Broglie wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: E2=(pc2)+(mc2)2 and K=Emc2.) (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

(a) What accelerating potential is needed to produce electrons of wavelength 5.00 nm? (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

Determine λm , the wavelength at the peak of the Planck distribution, and the corresponding frequency f, at these temperatures: (a) 3.00 K; (b) 300 K; (c) 3000 K.

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