Question

(a) Show that a potential difference of 1.02 x 10^6 V would be sufficient to give an electron a speed equal to twice the speed of light if Newtonian mechanics remained valid at high speeds. (b) What speed would an electron actually acquire in falling through a potential difference equal to 1.02 x 10^ V?

Answer 1

Answer on question #52486, Physics, Relativity

Question An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to $2.50\cdot 10^{-28}$ kg and that of the heavier fragment is $1.67\cdot 10^{-27}$ kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?

Solution Relativistic momentum is

$p=\frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

The momentums of fragments should be equal. Hence, momentum of heavier fragment $m_{2}$ is

$p=\frac{m_{1}0.893c}{\sqrt{1-0.893^{2}}}\approx 1.984m_{1}c$

From this we can find velocity of heavier fragment $v_{2}$

$p=\frac{m_{2}v_{2}}{\sqrt{1-\frac{v_{2}^{2}}{c^{2}}}}$

$p^{2}(1-\frac{v_{2}^{2}}{c^{2}})=m_{2}^{2}v_{2}^{2}$

$v_{2}^{2}(m_{2}^{2}+p^{2}/c^{2})=p^{2}$

$v_{2}=\sqrt{\frac{p^{2}}{(m_{2}+p^{2}/c^{2})}}=\frac{1.984m_{1}c}{\sqrt{m_{2}^{2}+1.984^{2}m_{1}^{2}}}\approx 0.285c$

http://www.AssignmentExpert.com/