Answer to Question #347489 in Physics for Abegail

Question

Compare the relativistic momentum of an electron to its classical momentum when it has a speed of (a) 0.88c in an accelerator used for treating cancer and (b)4.50 × 107 𝑚/𝑠 in the cathode ray tube of a TV set.

Answer 1

$$p_c=mv,\; p_r=\frac{mv}{\sqrt{1-(v/c)^2}}$$

(a)

$$p_c=9.1*10^{-31}*0.88*3*10^8\\=2.4*10^{-22}\:{\rm kg\cdot m/s}$$

$$p_r=\frac{mv}{\sqrt{1-(v/c)^2}}$$

$$=\frac{2.4*10^{-22}\:{\rm kg\cdot m/s} }{\sqrt{1-(0.88)^2}}=5.1*10^{-22}\:{\rm kg\cdot m/s}$$

(b)

$$p_c=9.1*10^{-31}*4.5*10^7\\=4.1*10^{-23}\:{\rm kg\cdot m/s}$$

$$p_r=\frac{mv}{\sqrt{1-(v/c)^2}}$$

$$=\frac{4.1*10^{-23}\:{\rm kg\cdot m/s} }{\sqrt{1-(4.5*10^7/3*10^8)^2}}=4.1*10^{-23}\:{\rm kg\cdot m/s}$$