A) \[\left( \frac{3c}{2} \right)\]
B) \[3\frac{c}{\sqrt{2}}\]
C) \[\frac{{{\left( 3c \right)}^{1/2}}}{2}\]
D) \[\frac{c\sqrt{3}}{2}\]
Correct Answer: D
Solution :
The relativistic kinetic energy of a particle of rest mass mo is given by \[K=(m-{{m}_{0}}){{c}^{2}}\] \[m=\frac{{{m}_{0}}}{\sqrt{1-({{v}^{2}}/{{c}^{2}})}},\] where m is the mass of the particle moving with velocity v. \[\therefore \] \[K=\left[ \frac{{{m}_{0}}}{\sqrt{1-({{v}^{2}}/{{c}^{2}})}}-{{m}_{0}} \right]{{c}^{2}}z\] According to problem, kinetic energy = rest energy \[\therefore \] \[\left[ \frac{{{m}_{0}}}{\sqrt{1-({{v}^{2}}/{{c}^{2}})}}-{{m}_{0}} \right]{{c}^{2}}={{m}_{0}}{{c}^{2}}\] or \[\frac{{{m}_{0}}{{c}^{2}}}{\sqrt{1-({{v}^{2}}/{{c}^{2}})}}=2{{m}_{0}}{{c}^{2}}\] or \[\frac{1}{1-({{v}^{2}}/{{c}^{2}})}=4\] or \[4{{v}^{2}}/{{c}^{2}}=3\] \[\therefore \] \[v=\frac{\sqrt{3}c}{2}\]
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