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$Cs\, Cl$ ના સામાન્ય સ્ફટકીના બંધારણમાં $Cs^+$ અને $Cl^-$ આયનો $bcc$ રચનામાં આકૃતિમાં દર્શાવ્યા પ્રમાણે ગોઠવાય છે. આઠ $Cs^+$ આયનોને લીધે $Cl^-$ આયન પર લાગતું ચોખ્ખું સ્થિતિ વિદ્યુત શાસ્ત્રનું બળ ....... છે.
$zero$
$\frac{1}{{4\pi \,\,{ \in _0}}}\,\frac{{16{e^2}}}{{3{a^2}}}$
$\frac{1}{{4\pi \,\,{ \in _0}}}\,\frac{{32{e^2}}}{{3{a^2}}}$
$\frac{1}{{4\pi \,\,{ \in _0}}}\,\frac{{4{e^2}}}{{3{a^2}}}$
Solution
Step $1:$ Net force on $Cl ^{-}$ion at the centre
From Figure, Position of $Cl ^{-}$is symmetric at the centre, and distances of all $Cs ^{+}$ ions from center is same, so magnitude of all attraction forces on $Cl ^{-}$will also be same.
Therefore, all $Cs ^{+}$ions will attract $Cl ^{-}$at center with equal forces in opposite directions. So, all their forces will cancel out. Hence Net force on $Cl ^{-}$ion will be $zero$.
Therefore correct answer is $D$
Step $2$: Detailed Explanation of forces on $Cl ^{-}$
The forces on $Cl ^{-}$will be get cancelled due to equal and opposite forces.
Forces due to $1$ and $7$ will cancel
Forces due to $2$ and $8$ will cancel
Forces due to $3$ and $5$ will cancel
Forces due to $4$ and $6$ will cancel
So, Net force on $Cl ^{-}$will be zero
