Solveeit Logo

Question

Physics Question on physical world

In the relation, p=αβeαZkθp = \frac{\alpha}{\beta} e^{-\frac{\alpha Z}{k\theta}} pp is pressure, ZZ is distance, kk is Boltzmann constant and θ\theta is the temperature. The dimensional formula of β\beta will be

A

[M0L2T0][ M^0 L^2 T^0 ]

B

[ML2T][ ML^2 T]

C

[ML0T1][ M L^0 T^{ - 1} ]

D

[M0L2T1][ M^0 L^2 T^{ - 1} ]

Answer

[M0L2T0][ M^0 L^2 T^0 ]

Explanation

Solution

[αZkθ]=[M0L0T0]\bigg [ \frac{ \alpha Z }{ k \theta } \bigg ] = [ M^0 L^0 T^0 ]
[α]=[kθZ][ \alpha ] = \bigg [ \frac{ k \theta }{ Z} \bigg ]
Further [ p ] = [αβ]\bigg [ \frac{ \alpha}{ \beta } \bigg ]
[β]=[αp]=[kθZp]\therefore [ \beta ] = \bigg [ \frac{ \alpha }{ p} \bigg ] = \bigg [ \frac{ k \theta }{ Z p } \bigg ]
Dimensions of k θ\theta are that to energy. Hence,
[β]=[ML2T2LML1T2][\beta ] = \bigg [ \frac{ ML^2 T^{ - 2}}{ LML^{ - 1} \, T^{ - 2} } \bigg ]
=[M0L2T0]= [ M^0 L^2 T^0 ]