એક દ્વીપરમાણ્વીય અણુમાં રહેલા બે પરમાણુઓ વચ્ચ | vedclass

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Question

$U\,\, = \,\,\frac{a}{{{x^{12}}}}\,\, - \,\,\frac{b}{{{x^6}}}$

$F\,\, = \,\, - \,\frac{{dU}}{{dx}}\,\, = \,\, + \;\,12\,\frac{a}{{{x^{13}}}}\,\, -

Vedclass · Accepted Answer

$\frac{{{b^2}}}{{4a}}$

Vedclass · Answer

$U\,\, = \,\,\frac{a}{{{x^{12}}}}\,\, - \,\,\frac{b}{{{x^6}}}$

$F\,\, = \,\, - \,\frac{{dU}}{{dx}}\,\, = \,\, + \;\,12\,\frac{a}{{{x^{13}}}}\,\, - \,\,\frac{{6b}}{{{x^7}}}\,\, = \,\,0\,\, \Rightarrow \,\,x\,\, = \,\,{\left( {\frac{{2a}}{b}} ight)^{1/6}}$

$U\,\,\left( {x\,\, = \,\,\infty \,} ight)\,\, = \,0$

${U_{equilibtium}}\,\, = \,\,\frac{a}{{{{\left( {\frac{{2a}}{b}} ight)}^2}}}\,\, - \,\,\frac{b}{{\left( {\frac{{2a}}{b}} ight)}}\,\, = \,\, - \frac{{{b^2}}}{{4a}}$

$	herefore \,\,U\,\,\left( {x\,\, = \,\,\infty } ight)\,\, - \,\,{U_{equilibrium}}\,\, = \,\,0\,\, - \,\,\left( {\frac{{ - {b^2}}}{{4a}}} ight)\,\, = \,\,\frac{{{b^2}}}{{4a}}$

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