If the velocity of light c, the constant of gravitation G and Planck's constant h be chosen as fundamental units, find the value of a gram, a centimeter and a second in terms of new units of mass, length and time respectively. Given `c = 3xx10^(10) cm s^(-1), G = 6.67xx10^(-8)` dyne `cm ^2 g^(-2), h = 6.6xx10^(-27)` erg sec.

Explanation

Here, `c =[LT^(-1)] = 3xx10^(10)cm s^(-1)`
`G = [M^(-1) L^3 T^(-2)] = 6.67xx 10^(-8) dyne cm^2 g^(-2)`
`h =[ML^2 T^(-1)] = 6.6xx10^(-27)erg. Sec.`
Now, `(hc)/(G) = (6.6xx10^(-27)xx3xx10^(10))/(6.67xx10^(-8))`
`(ML^2 T^(-1)xxLT^(-1))/(M^(-1) L^3 T^(-2)) = 2.9685xx10^(-9)`
`M^2 = 2.9685xx10^(-9)`
` M = sqrt(2.9685xx10^(-9)) = 0.5449xx 10^(-4)g`
`1g = (1)/(0.5448xx10^(-4))` new unit of mass `=1.8355 xx10^4` new unit of mass
Again , `(h)/(Mc^2) = (6.67xx10^(-27))/(0.5448xx10^(-4)xx(3xx10^(10))^2) = 1.303xx10^(-43)`
`(ML^2 T^(-1))/(ML^2T^(-2)) = 1.3603xx10^(-43)s or T = 1.3603xx10^(-43)s`
`:. 1s = (1)/(1.3603xx10^(-43)) = 0.735xx10^(43)` new unit of time
Again,`c xx T = 3xx10^(10)xx(1.3603xx10^(-43))`
`LT^(-1)xxT = L = 4.0809xx10^(-33)cm`
`1cm = (1)/(4.0809xx10^(-33)) = 0.2450xx10^(33)` new unit of length.