A length-scale (l) depends on the permittivit | vedclass

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(D) To determine the dimensionally correct expression,we analyze the dimensions of the given physical quantities:$[\varepsilon] = [M^{-1} L^{-3} T^4 A

Vedclass · Accepted Answer

$B, D$

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(D) To determine the dimensionally correct expression,we analyze the dimensions of the given physical quantities:$[\varepsilon] = [M^{-1} L^{-3} T^4 A^2]$$[k_B T] = [Energy] = [M L^2 T^{-2}]$$[q^2] = [A^2 T^2]$$[n] = [L^{-3}]$First,consider the expression $\frac{\varepsilon k_B T}{q^2}$:$\left[\frac{\varepsilon k_B T}{q^2}\right] = \frac{[M^{-1} L^{-3} T^4 A^2] [M L^2 T^{-2}]}{[A^2 T^2]} = [L^{-1}]$Now,check the options:For $(B)$: $l = \sqrt{\frac{\varepsilon k_B T}{n q^2}} \Rightarrow [l] = \sqrt{\frac{L^{-1}}{L^{-3}}} = \sqrt{L^2} = [L]$. This is dimensionally correct.For $(D)$: $l = \sqrt{\frac{q^2}{\varepsilon n^{1/3} k_B T}} \Rightarrow [l] = \sqrt{\frac{1}{L^{-1} (L^{-3})^{1/3}}} = \sqrt{\frac{1}{L^{-1} L^{-1}}} = \sqrt{L^2} = [L]$. This is dimensionally correct.Options $(A)$ and $(C)$ result in dimensions of $[L^{-2}]$ and $[L^{3/2}]$ respectively,which are incorrect for a length scale. Thus,$(B)$ and $(D)$ are correct.

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