1. Atomic potential (from Meyers & Chawla, Exercise 2.46) The potential energy o
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Question
1. Atomic potential (from Meyers & Chawla, Exercise 2.46) The potential energy of a Na' CT ion pair at the distance r is given by where q 1.6x10-1"C is the electric charge, 6,8.85x10N) s the permittivity of vacuum, and ,-12's the reference energy of two infintelyseparated ion If the equilibrium distance between the ions is ro0.276 nm a) Calculate the value of the constant B b) Calculate the total force between ions, and its attractive and repulsive portions, when r = 0.25 nm. Has a compressive or tensile force been applied to the ion pair to bring it to this configuration? c) Calculate the total force between ions, and its attractive and repulsive portions, when r 0.3 nm. Has a compressive or tensile force been applied to the ion pair to bring it to this configuration?Explanation / Answer
Use the relation 1 eV = 1.602*10-19 N.m; therefore, the reference energy Ur = 1.2 eV = (1.2 eV)*(1.602*10-19N.m/1 eV) = 1.9224*10-19 N.m
a) When the equilibrium separation between the ions is r0 = 0.276 nm = (0.276 nm)*(1 m/109 nm) = 2.76*10-10 m, then the potential energy U = 0. Plug in this condition in the equation and write
0 = Ur – q2/40r0 + B/r09
====> B/r09 = Ur - q2/40r0
Plug in values and obtain
B/r09 = 1.9924*10-19 N.m – (1.6*10-19C)2/4.(3.14).(8.85*10-12 C2.N-1.m-2).(2.76*10-10 m)
===> B/r09 = 1.9924*10-19 N.m – 8.3444*10-19 N.m = -6.352*10-19 N.m
===> B = (-6.352*10-19 N.m)*(2.76*10-10 m)9 = -5.9033*10-86*10-19 N.m10 -5.90*10-105 N.m10 (ans).
b) Use the value of B obtained from above; put r = 0.25 nm = (0.25 nm)*(1 m/10-9 m) = 2.5*10-10 m.
Plug in values and get
U = (1.9224*10-19 N.m) – (1.6*10-19C)2/4.(3.14).(8.85*10-12 C2.N-1.m-2).(2.5*10-10 m) + (-5.90*10-105N.m10)/(2.5*10-10 m)9 = (1.9224*10-19 N.m) – (9.2123*10-19 N.m) + (-5.90*10-105 N.m10)/(3.8147*10-87 m9) = -7.2199*10-19 N.m + (-1.5466*10-105+87) = -7.2199*10-19 N.m – 1.5466*10-18 = -(7.2199*10-19 + 1.5466*10-18) N.m = -2.26859*10-18 N.m -2.27*10-18 N.m (ans).
The force is negative and lower than the reference energy. The reference energy is the energy required to move the ions to infinity. Since the energy is negative, hence the ions are closer to each other and hence a compressive force is required (ans).
c) Use the value of B obtained from above; put r = 0.30 nm = (0.30 nm)*(1 m/10-9 m) = 3.0*10-10 m.
Plug in values and get
U = (1.9224*10-19 N.m) – (1.6*10-19C)2/4.(3.14).(8.85*10-12 C2.N-1.m-2).(3.0*10-10 m) + (-5.90*10-105N.m10)/(3.0*10-10 m)9 = (1.9224*10-19 N.m) – (7.6769*10-19 N.m) + (-5.90*10-105 N.m10)/(1.9683*10-86 m9) = -5.7545*10-19 N.m + (-2.9975*10-105+86) = -5.7545*10-19 N.m – 2.9975*10-19 = -(5.7545*10-19 + 2.9975*10-19) N.m = -8.752*10-19 N.m -8.75*10-19 N.m (ans).
The force is negative and lower than the reference energy. The reference energy is the energy required to move the ions to infinity. Since the energy is negative, hence the ions are closer to each other and hence a compressive force is required (ans).
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