What is the velocity o π- meson whose proper means life time is 2.5× 10⅞ second
We know that
τ=τ0(1−v2c2)\tau=\frac{\tau_0}{\sqrt{(1-\frac{v^2}{c^2}})}τ=(1−c2v2)τ0
τ=τ0×2.5×1078\tau=\tau_0\times2.5\times10^{\frac{7}{8}}τ=τ0×2.5×1087
τ0×2.5×1078=τ0(1−v2c2)\tau_0\times2.5\times10^{\frac{7}{8}}=\frac{\tau_0}{\sqrt{(1-\frac{v^2}{c^2}})}τ0×2.5×1087=(1−c2v2)τ0
Both sides take square
6.25×1074=11−v2c26.25\times10^{\frac{7}{4}}=\frac{1}{{1-\frac{v^2}{c^2}}}6.25×1047=1−c2v21
1−v2c2=.16×10−74{{1-\frac{v^2}{c^2}}}=.16\times10^{\frac{-7}{4}}1−c2v2=.16×104−7
1−v2c2=2.845×10−3{{1-\frac{v^2}{c^2}}}=2.845\times10^{-3}1−c2v2=2.845×10−3
v2c2=0.9971\frac{v^2}{c^2}=0.9971c2v2=0.9971
vc=0.9985\frac{v}{c}=0.9985cv=0.9985
v=0.9985cv=0.9985cv=0.9985c
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