Question 163: Units , Dimensions and Error Analysis

Question

The work done by a gas molecule in an isolated system is given by, $$W = \alpha {\beta ^2}{e^{ - {{{x^2}} \over {\alpha kT}}}}$$, where x is the displacement, k is the Boltzmann constant and T is the temperature. $$\alpha$$ and $$\beta$$ are constants. Then the dimensions of $$\beta$$ will be :

Accepted Answer

where, k is Boltzmann constant,
T is temperature and x is displacement.
We know that, $${{{x^2}} \over {\alpha kT}}$$ is a dimensionless quantity.
$$	herefore$$ $$\left[ {{{{x^2}} \over {\alpha kT}}} ight] = [{M^0}{L^0}{T^0}] \Rightarrow [\alpha ] = {{[{x^2}]} \over {[k][T]}}$$
$$ \Rightarrow [\alpha ] = {{[{L^2}]} \over {[k][T]}}$$ ..... (i)
Since, dimensions of k are
$$[k] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$$ ...... (ii)
Dimensions of temperature are
$$[T] = [K]$$ ..... (iii)
Substituting Eqs. (ii) and (iii) in Eq. (i), we get
$$[\alpha ] = {{[{L^2}]} \over {[{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}$$
$$[\alpha ] = [{M^{ - 1}}{T^2}]$$
According to dimensional analysis,
$$[W] = [\alpha {\beta ^2}]$$
$$ \Rightarrow [{\beta ^2}] = {{[W]} \over {[\alpha ]}}$$
$$ \Rightarrow [{\beta ^2}] = {{{M^1}{L^2}{T^{ - 2}}]} \over {[{M^{ - 1}}{T^2}]}} = [{M^2}{L^2}{T^{ - 4}}]$$
$$ \Rightarrow [\beta ] = [ML{T^{ - 2}}]$$

Answer

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

Answer

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

Answer

$$[ML{T^{ - 2}}]$$

Answer

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

Units , Dimensions and Error Analysis

The work done by a gas molecule in an isolated system is given by, $W = \alpha {\beta ^2}{e^{ - {{{x^2}} \over {\alpha kT}}}}, where x is the displacement, k is the Boltzmann constant and T is the temperature. \alpha and \beta are constants. Then the dimensions of \beta$ will be :

Choose the correct answer: