1.Units, Dimensions and Measurement
hard

બે પરમાણુઓ વચ્ચેની આંતરક્રિયાના બળને

$F=\alpha \beta \,\exp \,\left( { - \frac{{{x^2}}}{{\alpha kt}}} \right);$

વડે આપવામાં આવે છે, જ્યાં $x$ એ અંતર, $k$ બોલ્ટઝમેન અચળાંક અને $ T$ તાપમાન છે. તથા $\alpha$ અને $\beta$ એ અન્ય અચળાંકો છે. $\beta$ નું પરિમાણિક શું થાય?

A

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

B

$M^2LT^{-4}$

C

$MLT^{-2}$

D

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

(JEE MAIN-2019)

Solution

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

Standard 11
Physics

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