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
JEE MAIN 2019, Diffcult
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Power of $e$ should be dim ensionless.
So,$\left[ \lambda \right] = \left( {\alpha Tk} \right)$
$\Rightarrow \,{L^2} = \left[ \alpha \right]\left( {M{L^2}{T^{ - 2}}} \right)$
$\Rightarrow \left( \alpha \right) = \left( {{M^{ - 1}}{T^2}} \right)$
$\Rightarrow \,E = \frac{1}{2}KT$
$\Rightarrow \left( {M{L^2}{T^{ - 2}}} \right)\left( E \right) = \left[ {KT} \right]$
$\Rightarrow \left( {\alpha \beta } \right) = \left( F \right)$
$\Rightarrow \left( {{M^{ - 1}}{T^2}} \right)\left( \beta \right) = \left( {ML{T^{ - 2}}} \right)$
So,$\left[ \lambda \right] = \left( {\alpha Tk} \right)$
$\Rightarrow \,{L^2} = \left[ \alpha \right]\left( {M{L^2}{T^{ - 2}}} \right)$
$\Rightarrow \left( \alpha \right) = \left( {{M^{ - 1}}{T^2}} \right)$
$\Rightarrow \,E = \frac{1}{2}KT$
$\Rightarrow \left( {M{L^2}{T^{ - 2}}} \right)\left( E \right) = \left[ {KT} \right]$
$\Rightarrow \left( {\alpha \beta } \right) = \left( F \right)$
$\Rightarrow \left( {{M^{ - 1}}{T^2}} \right)\left( \beta \right) = \left( {ML{T^{ - 2}}} \right)$
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