Question

The dimensions of $\frac{e^{2}}{4\pi\varepsilon_{0}hc}$, where e, ε₀, h and c are electronic charge, electric permittivity, Planck’s constant and velocity of light in vacuum respectively is –

• A. $\lbrack M^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{0}\rbrack$
• B. $\lbrack M^{1}\text{ }\text{L}^{0}\text{ }\text{T}^{0}\rbrack$
• C. [M⁰ L¹ T⁰]
• D. $\lbrack M^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{1}\rbrack$

Answer 1

Answer: (A)

$\lbrack M^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{0}\rbrack$

Answer 2

$\frac{e^{2}}{4\pi\varepsilon_{0}hc}$

[$\frac{kq^{2}}{r^{2}}$ = F ⇒ ke² = Fr²

Energy E = $\frac{hc}{\lambda}$ ⇒ hc = Eλ]

$\frac{Fr^{2}}{hc} = \frac{Fr^{2}}{E\lambda}$= [M⁰L⁰T⁰]