JEE Advanced 2024 Paper 1 · Q01 · Dimensional Analysis

Question

A dimensionless quantity is constructed in terms of electronic charge $e$, permittivity of free space $\varepsilon_0$, Planck's constant $h$, and speed of light $c$. If the dimensionless quantity is written as $e^\alpha \varepsilon_0^\beta h^\gamma c^\delta$ and $n$ is a non-zero integer, then $(\alpha, \beta, \gamma, \delta)$ is given by

SolveKar AI · Accepted Answer

Dimensions: $[e] = AT$, $[\varepsilon_0] = M^{-1}L^{-3}T^4A^2$, $[h] = ML^2T^{-1}$, $[c] = LT^{-1}$. For $e^\alpha \varepsilon_0^\beta h^\gamma c^\delta$ to be dimensionless we need each of $M, L, T, A$ to have zero exponent: $M$: $-\beta + \gamma = 0$; $L$: $-3\beta + 2\gamma + \delta = 0$; $T$: $\alpha + 4\beta - \gamma - \delta = 0$; $A$: $\alpha + 2\beta = 0$. From the first, $\gamma = \beta$. From $A$: $\alpha = -2\beta$. Plug into $L$: $-3\beta + 2\beta + \delta = 0 \Rightarrow \delta = \beta$. Check $T$: $-2\beta + 4\beta - \beta - \beta = 0$ ✓. Setting $\beta = -n$ gives $(\alpha, \beta, \gamma, \delta) = (2n, -n, -n, -n)$. The famous example is the fine-structure constant $\alpha_{FS} = e^2/(4\pi \varepsilon_0 \hbar c)$, i.e. $n=1$.

JEE Advanced 2024 Paper 1 · Q01 · Dimensional Analysis

Solution