Question

calculate net potential energy of a na cl ion pair which formed by ionic bonding if it known that the ionic radius of na ion is 0.095nm and that of cl ion is 0.181nm the value of the electron charge is 1.6*10^-19 and of the permittivty of vacuum is 8.85*10^-12 C^2/(n-m^2).assume n=9 for nacl

Answer 1

force =  q / 4 ₀ r   

V (potential energy )= q₁ q₂ / 4 ₀ r² + b / r ⁿ r = r₊ + r_{- =} 0.095 +  0.181nm = 0.276 nm

= (1.6*10^{-19 )2}C / 4 * 3.14 * 8.85*10^-12 C²/(n-m²) (0.276)² nm + b / r ⁿ

= 0.3020 * 10 ^-8 *    + b / r ⁿ

force ( attractive ) rⁿ⁺¹/n = 0.3020 * 10 ^-8 (0.267 *10 -9)¹⁰ / 9 =

0.3020 * 10 ^-8 * 1.841 * 10 ^-25 /9=   0.3020 * 10 ^-8 * 0.2045 *   10 ^-25 = 0.0617 * 10^-33

b / r ⁿ   =  0.0617 * 10^-33/ (0.276*10^-9)⁹ =0.0617 * 10^-33/ 9.29 * 10^-87= 6.641 * 10^-54 Nm⁹

putting value in above equation we get

   0.3020 * 10 ^{-8 *}0.276*10^-9 + 6.641 * 10^-29 Nm⁹

=0.00833 * 10 ^-17 + 6.641 * 10^{-29 =} 8.33  * 10 ^-19 J