Question Number 41148 by Necxx last updated on 02/Aug/18
$${Calculate}\:{the}\:{speed}\:{of}\:{an}\:{electron} \\ $$$${whose}\:{kinetic}\:{energy}\:{is}\:{equal}\:{to} \\ $$$$\mathrm{2\%}\:{of}\:{its}\:{rest}\:{mass}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18
$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{100}}{m}_{\mathrm{0}} \\ $$$$\frac{{m}_{\mathrm{0}} }{\:\sqrt{\mathrm{1}−\frac{{v}^{\mathrm{2}} }{{c}^{\mathrm{2}} }}}\:\:×{v}^{\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{25}}{m}_{\mathrm{0}} \\ $$$$\mathrm{25}{v}^{\mathrm{2}} \:=\sqrt{\mathrm{1}−\frac{{v}^{\mathrm{2}} }{{c}^{\mathrm{2}} }}\: \\ $$$$\mathrm{25}{v}^{\mathrm{2}} \:\approx\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\frac{{v}^{\mathrm{2}} }{{c}^{\mathrm{2}} } \\ $$$$\mathrm{25}{v}^{\mathrm{2}\:} +\frac{\mathrm{1}}{\mathrm{2}}\frac{{v}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{50}{v}^{\mathrm{2}} {c}^{\mathrm{2}} +{v}^{\mathrm{2}} }{\mathrm{2}{c}^{\mathrm{2}} }=\mathrm{1} \\ $$$${v}^{\mathrm{2}} \left(\mathrm{50}{c}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{2}{c}^{\mathrm{2}} \\ $$$${v}=\sqrt{\frac{\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{1}+\mathrm{50}{c}^{\mathrm{2}} }}\:\:\:{c}={speed}\:{of}\:{light}... \\ $$
Commented by Necxx last updated on 03/Aug/18
$${Thank}\:{you}\:{so}\:{much}\:{sirTanmay}.. \\ $$$${please}\:{if}\:{I}\:{may}\:{ask}\:{how}\:{did}\:{you} \\ $$$${come}\:{about} \\ $$$$ \\ $$$${mv}^{\mathrm{2}} =\frac{{m}_{\mathrm{0}} }{\:\sqrt{\mathrm{1}\:−\frac{{v}^{\mathrm{2}} }{{c}^{\mathrm{2}} }}}\:\:\:\:?? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18
$${study}\:{relative}\:{theory}... \\ $$
Commented by Necxx last updated on 03/Aug/18
$${ok}\:{sir}.....{Thank}\:{you} \\ $$