The force of interaction between two atoms is | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\begin{array}{l}
Power\,of\,e\,should\,be\,\dim ensionless.\
So,\left[ \lambda  ight] = \left( {\alpha Tk} ight)\
 \Rightarrow \,\

Vedclass · Accepted Answer

$M^2LT^{-4}$

Vedclass · Answer

$\begin{array}{l}
Power\,of\,e\,should\,be\,\dim ensionless.\
So,\left[ \lambda  ight] = \left( {\alpha Tk} ight)\
 \Rightarrow \,\,\,\,\,\,\,\,{L^2} = \left[ \alpha  ight]\left( {M{L^2}{T^{ - 2}}} ight)\
 \Rightarrow \,\,\,\,\,\,\,\,\,\left( \alpha  ight) = \left( {{M^{ - 1}}{T^2}} ight)\
 \Rightarrow \,\,\,\,\,\,\,\,\,\,\,E = \frac{1}{2}KT\
 \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left( {M{L^2}{T^{ - 2}}} ight)\,\,;\,\,\left( E ight) = \left[ {KT} ight]\
 \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left( {\alpha \beta } ight) = \left( F ight)\
 \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\left( {{M^{ - 1}}{T^2}} ight)\left( \beta  ight) = \left( {ML{T^{ - 2}}} ight)
\end{array}$

The force of interaction between two atoms is given by $F\, = \,\alpha \beta \,\exp \,\left( { - \frac{{{x^2}}}{{\alpha kt}}} \right);$ where $x$ is the distance, $k$ is the Boltzmann constant and $T$ is temperature and $\alpha $ and $\beta $ are two constants. The dimension of $\beta $ is

Similar Questions

The foundations of dimensional analysis were laid down by

Obtain the relation between the units of some physical quantity in two different systems of units. Obtain the relation between the $MKS$ and $CGS$ unit of work.

In the relation $P = \frac{\alpha }{\beta }{e^{ - \frac{{\alpha Z}}{{k\theta }}}}$ $P$ is pressure, $Z$ is the distance, $k$ is Boltzmann constant and $\theta$ is the temperature. The dimensional formula of $\beta$ will be