In the relation \(P=\frac{\alpha}{\beta} e^{-\alpha z / K \theta}\) temperature. The dimensional formula of…
In the relation \(P=\frac{\alpha}{\beta} e^{-\alpha z / K \theta}\) temperature. The dimensional formula of \(\beta\) will be \(\left[M^{x} L^{y} T^{z}\right]\), then the value of \(x+y+z\) is
Where,
Solution
They've given the formula for Pressure as
\(P=\frac{\alpha}{\beta} \exp \left(-\frac{\alpha z}{k \theta}\right)\)
Where,
\(z\) is distance
\(k\) is Boltzmann's constant
\(\theta\) is temperature
Now, here we have to observe that the exponential does not have any units i.e.,
\(\frac{\alpha z}{k \theta}=\left[M^0 L^0 T^0\right]\)
We have the individual dimensional formulas for the quantities as
\(\begin{aligned} z & =\left[M^0 L^1 T^0\right] \\ k & =\left[M^1 L^2 T^{-2} K^{-1}\right] \\ \theta & =\left[M^0 L^0 T^0 K^1\right] \end{aligned}\)
Substituting these quantities in the above formula,
\(\begin{aligned} & \frac{\alpha z}{k \theta}=\left[M^0 L^0 T^0\right] \\ & \Rightarrow \frac{\alpha\left[M^0 L^1 T^0\right]}{\left[M^1 L^2 T^{-2} K^{-1}\right]\left[M^0 L^0 T^0 K^1\right]}=\left[M^0 L^0 T^0\right] \\ & \Rightarrow \frac{\alpha\left[M^0 L^1 T^0\right]}{\left[M^1 L^2 T^{-2} K^0\right]}=\left[M^0 L^0 T^0\right] \\ & \Rightarrow \alpha\left[M^{-1} L^{-1} T^2\right] \end{aligned}\)
Now, writing the dimension formula for the pressure we have,
\(\begin{aligned} & P=\frac{\alpha}{\beta} \exp \left(-\frac{\alpha z}{k \theta}\right) \\ & \Rightarrow[P]=\frac{[\alpha]}{[\beta]} \\ & \Rightarrow\left[M^1 L^{-1} T^{-2}\right]=\frac{\left[M^1 L^1 T^{-2}\right]}{[\beta]} \\ & \Rightarrow[\beta]=\frac{\left[M^1 L^1 T^{-2}\right]}{\left[M^1 L^{-1} T^{-2}\right]} \\ & \Rightarrow[\beta]=\left[M^0 L^2 T^0\right] \\ & \therefore[\beta]=\left[L^2\right] \end{aligned}\)
Asked in: JEE Mains - Units and Dimensions - Chapter Test