question_answer

Show that if e is electronic charge, h is Planck's constant, c is velocity of light and is electric permittivity of free space, then \[{{e}^{2}}/{{\in }_{0}}hc\] he is dimensionless.

Answer:

                We know that electrostatic potential energy between two electrons separated by distance r is \[U=\frac{ee}{4\pi {{\in }_{0}}r}\]             \[\therefore \] units of \[\frac{{{e}^{2}}}{{{\in }_{0}}}=U\times r=\left( M{{L}^{2}}{{T}^{-2}} \right)L\] Also. energy of a photon, \[E=h\upsilon =\frac{hc}{\lambda }\]\[\therefore \] units of \[hc=E\times \lambda =\left( M{{L}^{2}}{{T}^{-2}} \right)L\] Now \[\frac{{{e}^{2}}}{{{\in }_{0}}hc}=\frac{\left( M{{L}^{2}}{{T}^{-2}} \right)L}{\left( M{{L}^{2}}{{T}^{-2}} \right)L}=\left( {{M}^{0}}{{L}^{0}}{{T}^{0}} \right)\], i.e., the given quantity is dimensionless.