ke ^{2} / G m _{ e } m _{ p } ગુણોત્તર પરિમાણ | vedclass

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Question

The given ratio is $\frac{k e^{2}}{G m_{e} m_{p}} .$ Where, $G=$ Gravitational constant. Its unit is $N m ^{2} \,kg ^{-2}$

$m _{ c }$ and $m _{ p }

Vedclass · Accepted Answer

Vedclass · Answer

The given ratio is $\frac{k e^{2}}{G m_{e} m_{p}} .$ Where, $G=$ Gravitational constant. Its unit is $N m ^{2} \,kg ^{-2}$

$m _{ c }$ and $m _{ p }=$ Masses of electron and proton and their unit is kg.

$e =$ Electric charge. Its unit is $C$. $k=\frac{1}{4 \pi \varepsilon_{0}}$ and its unit is $N m ^{2} \,C ^{-2}$

Therefore, unit of the given ratio

$\frac{k e^{2}}{G m_{e} m_{p}}=\frac{\left[N m^{2} C^{-2}ight]\left[C^{-2}ight]}{\left[N\, m^{2}\, k g^{-2}ight][k g][k g]}$$=M^{0} L^{0} T^{0}$

Hence, the given ratio is dimensionless. $e=1.6 	imes 10^{-19} \,C$

$G=6.67 	imes 10^{-11}\, N m ^{2}\, kg ^{-2}$

$m _{ e }=9.1 	imes 10^{-31} \,kg$

$m _{ p }=1.66 	imes 10^{-27}\, kg$

Hence, the numerical value of the given ratio is

$\frac{k e^{2}}{G m_{e} m_{p}}=\frac{9 	imes 10^{9} 	imes\left(1.6 	imes 10^{-19}ight)^{2}}{6.67 	imes 10^{-11} 	imes 9.1 	imes 10^{-31} 	imes 1.67 	imes 10^{-27}}$$\approx 2.3 	imes 10^{39}$

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

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