Question

The energy released when a sodium ion and a chloride ion, originally at infinite distance are brought together to a distance of $2.76\mathring{A}$, closest distance in $NaCl$ crystal. Assume ions act as point charge and permitivity constant of the medium is $9\times {10}^{9}N{m}^2{C}^{-2}$ :

• A. $8.37\times {10}^{-19}J$
• B. $8.37\times {10}^{-10}J$
• C. $8.37\times {10}^{-20}J$
• D. $8.37\times {10}^{-25}J$

Answer 1

The correct option is A $8.37\times {10}^{-19}J$

When a charged particle is brought by infinite distance to a distance r in electric field of another charged particle, the work done is stored as potentail energy and given by:
$E=-k\cdot \frac{q_1\cdot q_2}{r}$
$E=-\frac{9\times {10}^9\times (1.602\times {10}^{-19})\times (-1.602\times {10}^{-19})}{2.76\times {10}^{-10}}$

Charge of sodium ion is positive and that of chloride ion is negative so:
$E=-8.37\times {10}^{-19}J$