Find out dimension of 1/4πε0 e2/hc where e : electronic charge, ε0 = permittivity of free space, h : plank constant

Question

Find out dimension of ${1\over4\pi\varepsilon_0}{e^2\over hc}$ where e : electronic charge, $\varepsilon_0$ = permittivity of free space, h : plank constant, c : speed of light

(1) M¹L¹T^–2C²

(2) M²L²T^–3C²

(3) M¹L¹T^–2C²

(4) Dimension less

Comments

Physics and measurement

Answer 1

Ans. (4) Dimension less

Sol. ${1\over4\pi\epsilon_0}{e^2\over hc}={Ke^2\times\lambda^2\over\lambda^2\times hc}={F\times\lambda\over E}={E\over E}$: dimension less