In the relation P = frac{alpha}{beta} e^{-fra | vedclass

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(A) In the given equation,the exponent $-\frac{\alpha Z}{k	heta}$ must be dimensionless.Therefore,the dimensions of $\frac{\alpha Z}{k	heta}$ are $[

Vedclass · Accepted Answer

$[M^0 L^2 T^0]$

Vedclass · Answer

(A) In the given equation,the exponent $-\frac{\alpha Z}{k	heta}$ must be dimensionless.Therefore,the dimensions of $\frac{\alpha Z}{k	heta}$ are $[M^0 L^0 T^0]$.Since $[k	heta]$ represents energy,its dimension is $[M L^2 T^{-2}]$.Thus,$[\alpha] = \frac{[k	heta]}{[Z]} = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$.Given $P = \frac{\alpha}{\beta}$,we have $[\beta] = \frac{[\alpha]}{[P]}$.The dimension of pressure $P$ is $[M L^{-1} T^{-2}]$.Therefore,$[\beta] = \frac{[M L T^{-2}]}{[M L^{-1} T^{-2}]} = [M^0 L^2 T^0]$.

In the relation $P = \frac{\alpha}{\beta} e^{-\frac{\alpha Z}{k\theta}}$,$P$ is pressure,$Z$ is distance,$k$ is Boltzmann constant,and $\theta$ is temperature. The dimensional formula of $\beta$ is:

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