Ncert Exercises & Solutions

Q 1.3) Check that the ratio ke²/G m_em_p is dimensionless. Lookup a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

The given ratio is

$\frac{k e^{2}}{G m_{e} m_{p}}$ Where G= Gravitational constant. Its unit is

$N m^{2} k g^{- 2}$

$m_{e} and m_{p}$ = Masses of electron and proton and their unit is kg.
e = Electric charge. Its unit is C.

$k = \frac{1}{4 π ε_{0}} and its unit is N m^{2} C^{- 2}$
Therefore, the unit of the given ratio

$\frac{k e^{2}}{G m_{e} m_{p}} = \frac{[N m^{2} C^{- 2}] [C^{- 2}]}{[N m^{2} k g^{- 2}] [k g] [k g]} = M^{0} L^{0} T^{0}$
Hence, the given ratio is dimensionless. WAR

$\begin{array}{l} e = 1.6 \times 10^{- 19} C \\ G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2} \\ m_{e} = 9.1 \times 10^{- 31} k g \\ m_{p} = 1.66 \times 10^{- 27} k g \end{array}$
Hence, the numerical value of the given ratio is

$\frac{k e^{2}}{G m_{e} m_{p}} = \frac{9 \times 10^{9} \times (1.6 \times 10^{- 19})^{2}}{6.67 \times 10^{- 11} \times 9.1 \times 10^{- 31} \times 1.67 \times 10^{- 27}} \approx 2.3 \times 10^{39}$
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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