A beam of particles strikes a wall containing \(2 \times 10^{29}\) atoms per \(\mathrm{m}^{3}\). Each atom behaves like a sphere of radius \(3 \times 10^{-15} \mathrm{~m}\). Find the thickness of wall that exactly half the particles will penetrate without scattering. What thickness would be needed to stop all but one particle in \(10^{6} ?\)

We know that each atom behaves like a sphere with a radius of \(3 \times 10^{-15}\) meters. To find the cross-sectional area of a single atom, we use the formula for the area of a circle: \(A = \pi r^2\). In this case, the cross-sectional area of a single atom is \[A = \pi (3 \times 10^{-15})^2 = 9 \pi \times 10^{-30} \mathrm{m}^2\]

We can find the probability of a single particle colliding with an atom by dividing the cross-sectional area of an atom by the total area in which the particle can move inside a cubic meter. Since there are \(2 \times 10^{29}\) atoms per cubic meter, the total cross-sectional area occupied by these atoms is \(2 \times 10^{29} \times 9\pi \times 10^{-30} \mathrm{m}^2\). Each particle would have a total area of \(1 \mathrm{m}^2\) to move. Therefore, the probability of a single particle colliding with an atom is, \[ P_1 = \frac{2 \times 10^{29} \times 9\pi \times 10^{-30} \mathrm{m}^2}{1 \mathrm{m}^2} = 18 \pi \times 10^{-1} \]

We are asked to find the thickness of the wall that would allow half of the particles to penetrate without scattering. To find this thickness, we must determine the probability of a particle not colliding with any atom. This probability is given by: \[ P_{not\_collide} = 1 - P_1 \] Since we want the probability of a particle not colliding with any atom in the entire thickness of the wall to be 50%, we can use the following formula: \[ P_{not\_collide}^{thickness} = 0.5 \] \[ (1 - P_1)^{thickness} = 0.5 \] Taking the natural logarithm of both sides, we get: \[thickness \cdot \ln(1 - P_1) = \ln(0.5)\] Solving for thickness, we get: \[ thickness = \frac{\ln(0.5)}{\ln(1-18 \pi \times 10^{-1})} \approx 4.6 \times 10^{-15} \mathrm{m} \]

We are asked to find the thickness of the wall that would allow only one particle out of one million to penetrate without scattering. Let the probability of a particle not colliding with any atom in the entire thickness of the wall for this case be \(P_{one}\). \[ P_{one} = \frac{1}{10^6} = 10^{-6} \] Using the same formula as in Step 3, we have: \[ (1 - P_1)^{thickness} = 10^{-6} \] Taking the natural logarithm of both sides, we get: \[ thickness \cdot \ln(1 - P_1) = \ln(10^{-6}) \] Solving for thickness, we get: \[ thickness = \frac{\ln(10^{-6})}{\ln(1-18 \pi \times 10^{-1})} \approx 1.4 \times 10^{-14} \mathrm{m} \] So, the thickness of the wall required to stop all but one particle in \(10^6\) is approximately \(1.4 \times 10^{-14} \mathrm{m}\).