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Units and Measurements — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements5 Marks
Question
A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?
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Answer

One relation consists of some fundamental constants that give the age of the Universe by: Where, t = Age of Universe e = Charge of electrons = $1.6 \times 10^{-19}C \in_0$ = Absolute permittivity
$m_p$ = Mass of protons = $1.67 \times 10^{-27}kg\ m_e$ = Mass of electrons = $9.1 \times 10^{-31}kg$ c = Speed of light = $3 \times 10^8m/s$ G = Universal gravitational constant = $6.67 \times 10^{11}Nm^2kg^{-2}$
Also, $\frac{1}{4\pi\in_0}=9\times10^9\text{Nm}^2/\text{C}^2$ Substituting these values in the equation, we get$\text{t}=\frac{(1.6\times10^{-19})^4\times(9\times10^9)^2}{(9.1\times10^{-31})\times1.67\times10^{-27}\times(3\times10^8)\times6.67\times10^{-11}}$
$=\frac{(1.6)^4\times81}{9.1\times1.67\times27\times6.67}\times10^{-76+18+62+27-24+11}\text{S}$
$=\frac{(1.6)^4\times81}{9.1\times1.67\times27\times6.67\times365\times24\times24\times3600}\times10^{-76+18+62+27-24+11}\text{years}$
$\approx6\times10^{-9}\times10^{18}\text{years}$
$=6\text{ billion years}$

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