Question

1. For a pair of ions M*-X, the attractive and repulsive energies EA and Er (in eV/pair), respectively, depend on the distance between the ions r (in nm), according to the following equations: 1.5 EA r ER 6 x 10-6 r9 a. Write down the expression for the total energy En. (2 marks) b. Take the derivative of Ex (4 marks) C. Calculate the equilibrium distance ro given that it is be distance at which en is minimum. (6 mark) d. Calculate the bond energy E. (2 marks) (Bonus question)

Answer 1

a)  

1.5 EN = 6 x 10-6- 29 + r

b)

DEN 54 x 10-6 1.5 r2 dr 1 10

c)  

Distancer, where En is minimun= 0.278 nm

d)  

Bond Energy (E) = 4.8 eV

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Explanation :

a)

En = EA + ER

1.5 EN = + 6x 10-6 .9 1

.

b)

1.5 EN = 6 x 10-6- 29 + r

DEN dr-9) -1.5 x dr-1) dr + 6 x 10-6 x dr dr

1 dEN dr = = -1.5 x -1) + 6 x 10-6 x (-9) 1 r 10

DEN = 1.5 r2 54 x 10-6 r10 dr

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c)

For maxima / minima

DEN = 0 dr

54x10-6 1T 15 x2 - 0 10

G 15 x2 - 54x10-6 ,.10

=> 1.5 r10 54 x 10-62

= r2( 1.5 78 54 x 10-6) = 0

Case 1 :

r2 = 0 =r=0 nm

Case 2 :

1.5 r8 – 54 x 10-6 = 0

54 x 10-6 1.5 18 = 0.278 nm ==

Now,

Double Derivative :

dPEN 3 = + 540 x 10-6 11 dr2 .3 r

Case 1 : When r = 0 nm

  

d’EN = -0+ = Not defined dr2

We don't have a definate maxima/minima at r = 0 nm.

But we can see that as r approaches zero, approches positive infinity, therefore it will be an indefinate maxima.

E N

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Case 2 : When r = 0.278 nm

d’EN 3 = + 540 x 10-6 (0.278) 11 dr2 (0.278) 3

dEN dr2 139.63 + + 704.53

d’EN = + 564.9 dr2

d’EN > 0 dr2

Since double derivative at r = 0.278 nm is positive, it is a minima

Therefore,

Distancer, where En is minimun= 0.278 nm

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d)

Bond Energy (E.) = En(r=r.)

1.5 = E. = + 6 x 10-6 re To

→ E = 1.5 0.278 6x10-6 (0.278) +

= E. -5.4 +0.6

= E. = -4.8 eV

Negative sign, because energy is released when bonds are formed.

.

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