Ncert Exercises & Solutions

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 the electron, the mass of a 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?

One relation consists of some fundamental constants that give the age of the Universe by:

$t = {(\frac{e^{2}}{4 π ε_{0}})}^{2} \times \frac{1}{m_{p} {m_{e}}^{2} c^{3} G}$

Where,

t = Age of Universe

e = Charge of electrons = 1.6 × $10^{- 19}$ C

$ε_{p}$ = Absolute permittivity

$m_{p}$ = Mass of protons = 1.67 × $10^{- 27}$ kg

$m_{e}$ = Mass of electrons = 9.1 × $10^{- 31}$ kg

c = Speed of light = 3 × $10^{8}$ m/s

G = Universal gravitational constant = 6.67 × $10^{11}$ $N m^{2} k g^{- 2}$

Also, $\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} / C^{2}$

Substituting these values in the equation, we get:

$t = \frac{{(1.6 \times 10^{- 19})}^{4} \times {(9 \times 10^{9})}^{2}}{1.67 \times 10^{- 27} \times {(9.1 \times 10^{- 31})}^{2} \times {(3 \times 10^{8})}^{3} \times 6.67 \times 10^{- 11}}$
$= \frac{{(1.6)}^{4} \times 81}{9.1 \times 9.1 \times 1.67 \times 27 \times 6.67} \times 10^{18}$
$= \frac{{(1.6)}^{4} \times 81}{9.1 \times 9.1 \times 1.67 \times 27 \times 6.67 \times 365 \times 24 \times 3600} \times 10^{18} years$
$\approx 6 \times 10^{- 9} \times 10^{18} years$
$= 6 billion years$

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