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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

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Physics and measurement

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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

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