question_answer

If the velocity of light c, the constant of graviation G and Planck's constant h be chosen as fundamentyal      units, find the value of a gram, a centimeter and a second in terms of new units of mass, length and time respectively. Given \[\text{c}=\text{3}\times \text{1}{{0}^{\text{1}0}}\text{cm }{{\text{s}}^{-\text{1}}},\text{ G}=\text{6}.\text{67}\times \text{1}{{0}^{-\text{8}}}\text{dyne c}{{\text{m}}^{\text{2}}}{{\text{g}}^{-\text{2}}},\text{ h}=\text{6}.\text{6}\times \text{1}{{0}^{-\text{27}}}\text{ergs}\].

Answer:

                Here, \[\text{c}=\left[ \text{L}{{\text{T}}^{-\text{1}}} \right]=\text{3}\times \text{1}{{0}^{\text{1}0}}\text{cm }{{\text{s}}^{-\text{1}}}\] \[\text{G}=\left[ {{\text{M}}^{-\text{1}}}{{\text{L}}^{\text{3}}}{{\text{T}}^{-\text{2}}} \right]=\text{6}.\text{67}\times \text{1}{{0}^{-\text{8}}}\text{dyne c}{{\text{m}}^{\text{2}}}{{\text{g}}^{-\text{2}}}\] \[\text{h}=\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{-\text{1}}} \right]=\text{ 6}.\text{6}\times \text{1}{{0}^{-\text{27}}}\text{erg}.\text{sec}.\]              Now,     \[\frac{hc}{G}=\frac{6.6\times {{10}^{-27}}\times 3\times {{10}^{10}}}{6.67\times {{10}^{-8}}}\]                 \[\frac{M{{L}^{2}}{{T}^{-1}}\times L{{T}^{-1}}}{{{M}^{-1}}{{L}^{3}}{{T}^{-2}}}=\text{ 2}.\text{9685}\times \text{1}{{0}^{-\text{9}}}\] \[{{\text{M}}^{\text{2}}}=\text{ 2}.\text{9685}\times \text{1}{{0}^{-\text{9}}}\]\[\text{M }=\sqrt{2.9685\times {{10}^{-9}}}\] \[\text{1g }=\frac{1}{0.5448\times {{10}^{-4}}}\]new unit of mass \[=\text{ 1}.\text{8355}\times \text{1}{{0}^{\text{4}}}\]new unit of mass Again,   \[\frac{h}{M{{C}^{2}}}=\frac{6.67\times {{10}^{-27}}}{0.5448\times {{10}^{-4}}\times {{(3\times {{10}^{10}})}^{2}}}=1.3603\times {{10}^{-43}}\] \[\frac{M{{L}^{2}}{{T}^{-1}}}{M{{L}^{2}}{{T}^{-2}}}=\text{ 1}.\text{36}0\text{3 }\times \text{ 1}{{0}^{-\text{43}}}\text{s}\] or    \[\text{T }=\text{1}.\text{36}0\text{3}\times \text{1}{{0}^{-\text{43}}}\text{s}\] \[\therefore \]   \[1s\,=\frac{1}{1.3603\times {{10}^{-43}}}=0.735\times {{10}^{43}}\]new unit of time Again, \[\text{c}\times \text{T}=\text{3}\times \text{1}{{0}^{\text{1}0}}\times \left( \text{1}.\text{36}0\text{3 }\times \text{1}{{0}^{-\text{43}}} \right)\] \[\left( \text{L}{{\text{T}}^{-\text{1}}} \right)\times \text{T}=\text{L}=\text{4}.0\text{8}0\text{9}\times \text{1}{{0}^{-\text{33}}}\text{cm}\] \[\text{1 cm }=\frac{1}{4.0809\times {{10}^{-33}}}=\text{ }0.\text{245}0\,\times \,{{10}^{33}}\]. new unit of length