Question

Asked Nov 25, 2019

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One model for the potential energy of a two-atom molecule, where the atoms are separated by a distance *r*, is * U*(*r*) = *U*_{0} [(*r*_{0} / *r*)^{16} - (*r*_{0} / *r*)^{9}]

where r0 = 0.70 nm and U0 = 7.0 eV. **Note: 1 eV = 1.6*10 ^{-19} J.**

[Force] = eV/nm

[Energy] = eV

[distance] = nm

r_eq = 7.6 *10^-1 nm

- If the distance between the atoms increases from equilibrium by r1 = 0.20 nm, then what is the force from one atom on the other associated with this potential energy? (Enter your answer as postive if they repel each other, and negative if they attract). F_r (r_eq +r_1) = ?
- The atoms are oscillating back and forth. The maximum separation of the atoms is r2 = 2.2 nm. What is the kinetic energy of the atoms when they are separated by the equilibrium distance? K(r_eq) = ?

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Step 1

Given data:

r_{o }= 0.70 nm = 0.70 x 10^{-9} m

Increased distance, r_{1} = 0.20 nm

r_{eq }= 7.6 x 10^{-1} nm = 0.76 nm

r = r_{eq} + r_{1} = 0.76 + 0.20 = 0.96 nm

U_{0} = 7.0 eV

1 eV = 1.6 x 10^{-19} J

Step 2

Formula used:

Step 3

Therefore...

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