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JEE Advanced 2024 · Paper 1, Physics question 1

A dimensionless quantity is constructed in terms of electronic charge ee, permittivity of free space ε0\varepsilon_0, Planck's constant hh, and speed of light cc. If the dimensionless quantity is written as eαε0βhγcδe^{\alpha}\varepsilon_0^{\beta}h^{\gamma}c^{\delta} and nn is a non-zero integer, then (α,β,γ,δ)(\alpha,\beta,\gamma,\delta) is given by

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Answer and worked solution

Official answer: A

The fine-structure constant e24πε0ℏc\dfrac{e^2}{4\pi\varepsilon_0\hbar c} is dimensionless, so e2ε0−1h−1c−1e^2\varepsilon_0^{-1}h^{-1}c^{-1} is dimensionless, and so is any power nn of it:
(α,β,γ,δ)=(2n,−n,−n,−n).(\alpha,\beta,\gamma,\delta) = (2n, -n, -n, -n).

Topics:Dimensional analysis
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