Question

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 $$v=\frac{c}{\sqrt{1}+(mc\lambda /h{)}^2}$$ (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 $v=(1-\mathrm{\Delta })c$ and express the speed of the electron in terms of rather than v. Find an expression for $\delta$ valid when $\lambda \ll hmc$. [$Hint:$ 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 $v=(1-\mathrm{\Delta })c$, and state the value of $\mathrm{\Delta }$

Short answer

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

Step-by-step solution

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$$\lambda =\frac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}$$.

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.

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:$$\frac{c}{\sqrt{1+{\left(\frac{mc\lambda }{h}\right)}^2}}\approx c(1-\frac{1}{2}{\left(\frac{mc\lambda }{h}\right)}^2)$$.Thus, $v=(1-\mathrm{\Delta })c$ and $\mathrm{\Delta }=\frac{1}{2}{\left(\frac{mc\lambda }{h}\right)}^2$.

Calculate Electron Speed for Given Wavelength

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

Key concepts

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 $(1-\mathrm{\Delta })c$.
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.