Question

Check that the ratio \(\frac{ke^2}{ Gm_e m_p}\) is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Answer 1

The given ratio is \(\frac{ke^2}{ Gm_e m_p}\). Where,
G = Gravitational constant. Its unit is \(Nm^2kg^{−2}\)
\(m_e\) and \(m_p\) = Masses of electron and proton and their unit is kg.
e = Electric charge. Its unit is C.
\(k = \frac{1}{4πε_0}\) and its unit is \(Nm^2C^{−2}\)
Therefore, unit of the given ratio
                         \(\frac{ke^2}{ Gm_e m_p}\) = \(\frac{[Nm^2C^{−2}][C^{-2}]}{[Nm^2kg^{−2}][Kg][Kg]}  = M^0L^0 T^0\)  
Hence, the given ratio is dimensionless.
e = \(1.6 × 10^{−19}C\)
G = \(6.67 × 10−11Nm^2kg^{−2}\)
\(m_e=\) \(9.1 × 10^{−31} kg\)
\(m_p = 1.66 × 10^{−27} kg\)
Hence, the numerical value of the given ratio is
\(\frac{ke^2}{ Gm_e m_p}\) \(= \frac{9 × 10^9 × (1.6 ×  10^{-19})^{2}} {6.67 ×  10^{-11} ×  9.1 ×  10^{ -31}× 1.67 ×  10^{ -27 }}≈ 2.3 × 10^{39}\)
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.