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One of the elementary bosons that can mediate electroweak interactions is the \(Z^{0}\) boson, having the mass of \(91.1876 \mathrm{GeV} / \mathrm{c}^{2}\). Find the order of magnitude of the range of the electroweak interaction.

Short Answer

Expert verified
Answer: The order of magnitude of the range of the electroweak interaction mediated by the \(Z^{0}\) boson is approximately \(10^{-18}\,m\).

Step by step solution

01

Calculate the energy of the \(Z^{0}\) boson

First, we need to calculate the energy of the \(Z^{0}\) boson using its mass and the speed of light (c). The energy is given by the formula: $$ E = mc^2 $$ where \(m\) is the mass of the particle, and \(c\) is the speed of light. For the \(Z^{0}\) boson, the mass \(m = 91.1876\,\frac{GeV}{c^2}\), and \(c = 3 \times 10^8\,\frac{m}{s}\). Let's plug these values into the formula: $$ E = 91.1876\,\frac{GeV}{c^2} \times c^2 = 91.1876\,GeV $$ So, the energy of the \(Z^{0}\) boson is \(91.1876\,GeV\).
02

Calculate the minimum uncertainty in momentum using uncertainty principle

Now, we can apply the uncertainty principle to estimate the minimum uncertainty in momentum (\(\Delta p\)) for the \(Z^{0}\) boson. According to the uncertainty principle: $$ \Delta x \Delta p \approx \hbar $$ where \(\Delta x\) is the range of the interaction, \(\Delta p\) is the uncertainty in momentum, and \(\hbar\) is the reduced Planck constant. Since we are concerned with the order of magnitude of the range, we will consider the uncertainty principle in the following approximate form: $$ \Delta x \approx \frac{\hbar}{\Delta p} $$ We can calculate the minimum uncertainty in momentum by dividing the energy \(E\) by the speed of light \(c\). This would give us: $$ \Delta p \approx \frac{E}{c} $$ Substituting the values, we have: $$ \Delta p \approx \frac{91.1876\,GeV}{3 \times 10^8\,\frac{m}{s}} $$
03

Calculate the order of magnitude of the range of the interaction

Now, using the minimum uncertainty in momentum, we can find the order of magnitude of the range of the interaction (\(\Delta x\)): $$ \Delta x \approx \frac{\hbar}{\Delta p} $$ Substituting the values, we have: $$ \Delta x \approx \frac{6.582 \times 10^{-16}\,\frac{eV \cdot m}{s}}{\frac{91.1876\,GeV}{3 \times 10^8\,\frac{m}{s}}} $$ We can convert the energy of the boson to eV by multiplying with \(10^9\): $$ \Delta x \approx \frac{6.582 \times 10^{-16}\,\frac{eV \cdot m}{s}}{\frac{91.1876 \times 10^9\,eV}{3 \times 10^8\,\frac{m}{s}}} $$ After calculations, we get: $$ \Delta x \approx 2.43 \times 10^{-18}\,m $$ So, the order of magnitude of the range of the electroweak interaction mediated by the \(Z^{0}\) boson is approximately \(10^{-18}\,m\).

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