High-speed electrons are used to probe the interior...

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, $p=m v / \sqrt{1-v^{2} / c^{2}}$ (a) Show that the speed of an electron that has de Broglie wavelength $\lambda$ is $$v=\frac{c}{\sqrt{1+(m c \lambda / h)^{2}}}$$ (b) The quantity $h / m c$ equals $2.426 \times 10^{-12} \mathrm{m}$ . (As we saw in Section $38.3,$ this same quantity appears in Eq. $(38.7),$ the expression for Compton sattering of photons by electrons.) If $\lambda$ is small compared to $h / m c,$ 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 $\Delta$ rather than $v .$ Find an expression for $\Delta$ valid when $\lambda \ll h / m c$ [Hint: Use the binomial expansion $(1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots$ valid for the case $|z|<1.1(\mathrm{c})$ How fast must an electron move for its de Broglie wavelength to be $1.00 \times 10^{-15} \mathrm{m},$ comparable to the size of a proton? Express your answer in the form $v=(1-\Delta) c,$ and state the value of $\Delta .$

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, $p=m v / \sqrt{1-v^{2} / c^{2}}$ (a) Show that the speed of an
electron that has de Broglie wavelength $\lambda$ is
$$v=\frac{c}{\sqrt{1+(m c \lambda / h)^{2}}}$$
(b) The quantity $h / m c$ equals $2.426 \times 10^{-12} \mathrm{m}$ . (As we saw in
Section $38.3,$ this same quantity appears in Eq. $(38.7),$ the expression for Compton sattering of photons by electrons.) If $\lambda$ is small
compared to $h / m c,$ 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 $\Delta$ rather than $v .$ Find an expression for $\Delta$ valid when $\lambda \ll h / m c$ [Hint: Use the binomial expansion $(1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots$ valid for the case $|z|<1.1(\mathrm{c})$ How fast must an electron move for its de Broglie
wavelength to be $1.00 \times 10^{-15} \mathrm{m},$ comparable to the size of a proton? Express your answer in the form $v=(1-\Delta) c,$ and state the
value of $\Delta .$

Key Concepts

De Broglie Wavelength

This concept establishes the wave-particle duality by relating a particle's momentum to its wavelength through the equation ? = h/p. It is fundamental to quantum mechanics and is used to describe how particles such as electrons exhibit wave-like behavior. In high-speed regimes, the de Broglie wavelength is key to understanding interactions at subatomic scales.

Relativistic Momentum

Relativistic momentum modifies the classical momentum expression to account for effects as a particle's speed approaches the speed of light. It is given by p = m v / ?(1 - v²/c²), where the denominator (the Lorentz factor) introduces the necessary corrections for time dilation and length contraction. This adjustment is crucial when dealing with high-speed particles like electrons in nuclear probes.

Binomial Expansion

The binomial expansion is a mathematical tool used to approximate expressions when a term in the expression is much smaller than unity. In the context of relativistic electron speeds, it is applied to simplify complex expressions (such as the square root in the Lorentz factor) when dealing with small deviations from the speed of light, allowing for easier analytical approximations.

Characteristic Length Scale (Compton Wavelength)

The quantity h/(mc), known as the Compton wavelength, represents a fundamental length scale in quantum physics. It appears in phenomena such as Compton scattering and serves as a benchmark for when quantum and relativistic effects become important. Comparing the de Broglie wavelength to the Compton wavelength helps determine the regime (classical or relativistic) in which an electron is operating.

Transcript

00:01 In this question, we are basically looking at a relativistic situation for an electron and you want to find the velocity with respect to a relativistic situation.

00:17 So in this case, the momentum, we will have to use the relativistic momentum, gamma times mv, where gamma is actually equals to square root of, right, 1 over square root of 1 minus v square over c square.

00:35 What we're going to do is we're going to bring this square to the left -hand side and we're going to square both sides.

00:48 This is what we get.

00:50 Then we're going to bring the term with v square over to the right -hand side to bm square plus b square over the c square times v square.

01:12 And if you were to rearrange this you can get b square.

01:22 This is the expression and now we can take the square root of this entire expression.

01:39 So now to simplify this bit and to make it into the expression that we need, right, from the question, we're going to multiply c over c to the numerator and denominator, and we're also going to divide the numerator and denominator by p.

02:04 1 over p.

02:11 So what we will get from here is that v equals to c over like we have m square c square now and we have to divide by p square plus plus one.

02:40 Now we substitute in p, what is p? right, the de -bloglyce wavelength formula.

02:45 P is equal to h over lambda right this still holds as i mentioned earlier in the question that it still holds for relativistic equations all right and so therefore this would be equals to c over square root substituting in p equals to h over lambda we get c lambda over h square plus one which is the expression that is stated in the question.

03:25 Now to be need to do a bit of binomial expansion.

03:34 We're going to simplify this formula a bit.

03:38 We bring the fraction up, right? so this is just c multiplied by 1 plus h square to the power half.

03:59 Now given that, sorry this is negative half right because it's from the denominator...

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