General Physics Ques 4
- In the relation, $p=\frac{\alpha}{\beta} e^{-\frac{\alpha Z}{k \theta}}$ $p$ is pressure, $Z$ is distance, $k$ is Boltzmann’s constant and $\theta$ is the temperature. The dimensional formula of $\beta$ will be
(2004, 2M)
(a) $\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^0\right]$
(b) $\left[\mathrm{ML}^2 \mathrm{~T}\right]$
(c) $\left[\mathrm{ML}^0 \mathrm{~T}^{-1}\right]$
(d) $\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-1}\right]$
Show Answer
Answer:
Correct Answer: 4.( a )
Solution:
$ \begin{array}{ll} & {\left[\frac{\alpha Z}{k \theta}\right]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]} \\ & {[\alpha]=\left[\frac{k \theta}{Z}\right]} \\ \text { Further } & {[p]=\left[\frac{\alpha}{\beta}\right]} \\ \therefore \quad & {[\beta]=\left[\frac{\alpha}{p}\right]=\left[\frac{k \theta}{Z p}\right]} \end{array} $
Dimensions of $k \theta$ are that to energy. Hence,
$ \begin{aligned} {[\beta] } & =\left[\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{LML}^{-1} \mathrm{~T}^{-2}}\right] \\ & =\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^0\right] \end{aligned} $
BETA
