Number of particles is given by n=-D (n2-n1/x2-x1) crossing a unit area perpendicular to X-axis is unit time, where n1 and n2 are number of particles per unit volume for the value of x meant to x2 and x1. Find dimensions of D called as diffusion constant.

Question

Number of particles is given by n=-D (n2-n1/x2-x1) crossing a unit area perpendicular to X-axis is unit time, where n1 and n2 are number of particles per unit volume for the value of x meant to x2 and x1. Find dimensions of D called as diffusion constant. (A) [ M 0 LT 2] (B) [M0 L2 T-4] (C) [ M 0 LT -3] (D) [ M 0 L 2 T -1]

Tardigrade Admin · Accepted Answer

Given, number of particle passing from unit area in unit time

= n

= \frac{Number of particle}{A \times t} = \frac{[ M ^{0} L ^{0} T ^{0} ]}{[ L ^{2} ] [ T ]} = [L^{- 2} T^{- 1}]

[n_{1}] = [n_{2}] =

Number of particle in unit volume

= [L^{- 3}]

Now, from the given formula

[D] = \frac{[ n ] [ x _{2} - x _{1} ]}{[ n _{2} - n _{1} ]}

= \frac{[ L ^{- 2} T ^{- 1} ] [ L ]}{[ L ^{- 3} ]}

= [L^{2} T^{- 1}]

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

$[M^{0} L^{2} T^{- 4}]$

$[M^{0} L T^{- 3}]$

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

Q. Number of particles is given by n=−Dx2​−x1​n2​−n1​​ crossing a unit area perpendicular to X-axis is unit time, where n1​ and n2​ are number of particles per unit volume for the value of x meant to x2​ and x1​. Find dimensions of D called as diffusion constant.

[M0LT2]

[M0L2T−4]

[M0LT−3]

[M0L2T−1]

Given, number of particle passing from unit area in unit time =n =A×t Number of particle ​=[L2][T][M0L0T0]​=[L−2T−1] [n1​]=[n2​]= Number of particle in unit volume =[L−3] Now, from the given formula [D]=[n2​−n1​][n][x2​−x1​]​ =[L−3][L−2T−1][L]​ =[L2T−1]

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

$[M^{0} L^{2} T^{- 4}]$

$[M^{0} L T^{- 3}]$

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