Question

Some particle detectors measure the total number of particles integrated over part of a sphere of radius \(R,\) where the target is at the center of the sphere. Assuming symmetry about the axis of the incoming particle beam, use the Rutherford scattering formula to obtain the total number of particles detected in an an interval of width \(d \theta\) as a function of the scattering angle, \(\theta .\)

Short answer

Q: Using the Rutherford scattering formula, calculate the total number of particles detected in an interval of width \(d\theta\). A: The total number of particles detected in an interval of width \(d\theta\), as a function of the scattering angle, can be calculated using the following formula: \[ dN = F \frac{Z_1^2Z_2^2e^4}{4\pi\epsilon_0^2 E^2} \cdot \frac{\cos(\frac{\theta}{2})d\theta}{\sin^3(\frac{\theta}{2})} \] where \(F\) is the incoming particle flux, \(Z_1\) and \(Z_2\) are the atomic numbers of the projectile and target nuclei, respectively, \(e\) is the electron charge, \(E\) is the kinetic energy of the projectile, and \(\epsilon_0\) is the vacuum permittivity.

Step-by-step solution

Write down the Rutherford scattering formula

The Rutherford scattering formula for the differential cross-section in terms of the scattering angle \(\theta\) is given by: \[ \frac{d\sigma}{d\Omega} = \frac{1}{4\pi\epsilon_0^2} \frac{Z_1^2Z_2^2e^4}{4E^2 \cdot sin^4(\frac{\theta}{2})} \] where \(Z_1\) and \(Z_2\) are the atomic numbers of the projectile and target nuclei, respectively, \(e\) is the electron charge, \(E\) is the kinetic energy of the projectile, and \(\epsilon_0\) is the vacuum permittivity.

Calculate the total number of particles detected

Now, we need to find the total number of particles detected in an interval \(d\theta\). We can do this by integrating the differential cross-section over the solid angle \(d\Omega\) corresponding to the interval \(d\theta\). The solid angle can be expressed as: \[ d\Omega = 2\pi\sin\theta d\theta \] Next, we need to multiply the differential cross-section by the incoming particle flux, \(F\), and the total solid angle \(d\Omega\) to get the total number of particles detected in the interval \(d\theta\). This can be expressed as: \[ dN = F \cdot \frac{d\sigma}{d\Omega} \cdot d\Omega \] Substituting the expressions for the Rutherford scattering formula and the solid angle, we have: \[ dN = F \left( \frac{1}{4\pi\epsilon_0^2} \frac{Z_1^2Z_2^2e^4}{4E^2 \cdot sin^4(\frac{\theta}{2})} \right)(2\pi\sin\theta d\theta) \]

Simplify the expression for the total number of particles detected

Now, we can simplify the expression for \(dN\) by canceling out common factors and combining terms: \[ dN = F \frac{Z_1^2Z_2^2e^4}{4\pi\epsilon_0^2 2E^2} \frac{\sin(\theta)d\theta}{\sin^4(\frac{\theta}{2})} \] Using the trigonometric identity \(\sin(2x) = 2\sin(x)\cos(x)\), we can rewrite the expression for \(\sin(\theta) = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})\), and then simplify further: \[ dN = F \frac{Z_1^2Z_2^2e^4}{4\pi\epsilon_0^2 2E^2} \frac{2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})d\theta}{\sin^4(\frac{\theta}{2})} \] Finally, we get the expression for the total number of particles detected in an interval of width \(d\theta\) as a function of the scattering angle: \[ dN = F \frac{Z_1^2Z_2^2e^4}{4\pi\epsilon_0^2 E^2} \cdot \frac{\cos(\frac{\theta}{2})d\theta}{\sin^3(\frac{\theta}{2})} \] This formula gives the total number of particles detected in an interval width \(d\theta\) for the given detector geometry and Rutherford scattering process.