JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 18)
The force of interaction between two atoms is given by F = $$\alpha $$$$\beta $$exp $$\left( { - {{{x^2}} \over {\alpha kt}}} \right)$$; 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 :
M2L2T$$-$$2
M2LT$$-$$4
MLT$$-$$4
M0L2LT$$-$$4
Explanation
$$F = \alpha \beta {e^{\left( {{{ - {x^2}} \over {\alpha KT}}} \right)}}$$
$$\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}$$
$${{{L^2}} \over {\left[ \alpha \right]M{L^2}{T^{ - 2}}}}$$ $$=$$ $${M^o}{L^o}{T^o}$$
$$ \Rightarrow $$ $$\left[ \alpha \right] = {M^{ - 1}}{T^2}$$
$$\left[ F \right] = \left[ \alpha \right]\left[ \beta \right]$$
MLT$$-$$2 = M$$-$$1T2[$$\beta $$]
$$ \Rightarrow $$ [$$\beta $$] = M2LT$$-$$4
$$\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}$$
$${{{L^2}} \over {\left[ \alpha \right]M{L^2}{T^{ - 2}}}}$$ $$=$$ $${M^o}{L^o}{T^o}$$
$$ \Rightarrow $$ $$\left[ \alpha \right] = {M^{ - 1}}{T^2}$$
$$\left[ F \right] = \left[ \alpha \right]\left[ \beta \right]$$
MLT$$-$$2 = M$$-$$1T2[$$\beta $$]
$$ \Rightarrow $$ [$$\beta $$] = M2LT$$-$$4
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