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Question Number 98003 by M±th+et+s last updated on 11/Jun/20 | ||
$${prove}\:{that}\: \\ $$$${E}={mc}^{\mathrm{2}} \:\: \\ $$ | ||
Answered by Rio Michael last updated on 11/Jun/20 | ||
$$\mathrm{Let}\:\mathrm{me}\:\mathrm{begin}\:\mathrm{with}\:\mathrm{Plank}'\mathrm{s}\:\mathrm{equation} \\ $$$$\:{E}\:=\:{hf}\: \\ $$$$\Rightarrow\:{E}\:=\:{h}\left(\frac{{c}}{\lambda}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}\:=\:{f}\lambda \\ $$$$\:\Rightarrow\:{E}\:=\:\frac{{hc}}{\lambda} \\ $$$$\mathrm{also}\:{E}\:=\:{pc}\:\mathrm{as}\:\mathrm{we}\:\mathrm{know}\:\mathrm{momentum}\:=\:\frac{{h}}{\lambda} \\ $$$$\Rightarrow\:{E}\:=\:{mc}\left({c}\right)\:,\:\mathrm{momentum}\:=\:{mv}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:{v}\:=\:{c} \\ $$$$\Rightarrow\:{E}\:=\:{mc}^{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{me}\:\mathrm{use}\:\mathrm{other}\:\mathrm{methods}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{this} \\ $$ | ||
Answered by Rio Michael last updated on 11/Jun/20 | ||