c) A beam of electrons moving in the positive x-direction impacts a target a vacuum chamber. If 2times 10^14 electrons traveling at a speed of 1times 10^7m strike the target during each brief pulse lasting 5times 10^-8s , Determine: (i). Porce exerted on the target during the pulse. (5marks) (iii). Pressure is exerted on the beam spot , which has radiue 4mm.

To determine the force exerted on the target during the pulse, we can use the concept of momentum transfer. The force exerted on the target is equal to the rate of change of momentum of the electrons striking the target.<br /><br />(i) Force exerted on the target during the pulse:<br /><br />Given:<br />- Number of electrons, N = $2\times 10^{14}$<br />- Speed of electrons, v = $1\times 10^{7}m/s$<br />- Duration of pulse, t = $5\times 10^{-8}s$<br /><br />The force exerted on the target can be calculated using the formula:<br /><br />Force = $\frac{N \cdot m \cdot v}{t}$<br /><br />where m is the mass of an electron.<br /><br />The mass of an electron is approximately $9.11 \times 10^{-31}$ kg.<br /><br />Substituting the given values, we get:<br /><br />Force = $\frac{2\times 10^{14} \cdot 9.11 \times 10^{-31} \cdot 1\times 10^{7}}{5\times 10^{-8}}$<br /><br />Simplifying the expression, we get:<br /><br />Force = $3.64 \times 10^{-12}$ N<br /><br />Therefore, the force exerted on the target during the pulse is $3.64 \times 10^{-12}$ N.<br /><br />(iii) Pressure exerted on the beam spot:<br /><br />Given:<br />- Radius of the beam spot, r = 4 mm = $4 \times 10^{-3}$ m<br /><br />The pressure exerted on the beam spot can be calculated using the formula:<br /><br />Pressure = $\frac{Force}{Area}$<br /><br />The area of the beam spot can be calculated using the formula for the area of a circle:<br /><br />Area = $\pi \cdot r^2$<br /><br />Substituting the given values, we get:<br /><br />Area = $\pi \cdot (4 \times 10^{-3})^2$<br /><br />Area = $50.27 \times 10^{-6}$ m²<br /><br />Now, substituting the force calculated earlier, we get:<br /><br />Pressure = $\frac{3.64 \times 10^{-12}}{50.27 \times 10^{-6}}$<br /><br />Simplifying the expression, we get:<br /><br />Pressure = $7.23 \times 10^{-7}$ Pa<br /><br />Therefore, the pressure exerted on the beam spot is $7.23 \times 10^{-7}$ Pa.