A dimensionless quantity is constructed in terms of electronic charge $e$, permittivity of free space…

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
  1. $(2 n,-n,-n,-n)$
  2. $(n,-n,-2 n,-n)$
  3. $(n,-n,-n,-2 n)$
  4. $(2 n,-n,-2 n,-2 n)$

Solution

For the quantity to be dimensionless $\begin{aligned} & \mathrm{e}^\alpha \varepsilon_0^\beta \mathrm{h}^\gamma \mathrm{c}^{\mathrm{d}}=\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}^0 \\ & \Rightarrow(\mathrm{AT})^\alpha\left(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right)^\beta\left(\mathrm{ML}^2 \mathrm{~T}^{-1}\right)^\gamma\left(\mathrm{LT}^{-1}\right)^\delta=\mathrm{A}^0 \mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \\ & \therefore \alpha+2 \beta=0, \alpha+4 \beta-\gamma-\delta=0,-\beta+\gamma=0 ~\&~-3 \beta+2 \gamma+\delta=0 \\ & \therefore \alpha=-2 \beta, \beta=\gamma ~\&~ \gamma=\delta\end{aligned}$ $\therefore$ Option (1) satisfies the given condition

Asked in: JEE Advanced 2024 (Paper 1)

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