Answer to Question #108124 in Physical Chemistry for Roopali Jha

Question

The atomic packing factor of iron decreases by what % when it is cooled from 1000 to 900 o C (iron is bcc at lower and fcc at higher temperature) Calculate the number of Co ions released (as m×10n ) in a year from the head (14 mm dia) of a hip joint prosthesis made of an alloy containing 45% Co). The wear rate of the head is 0.11 mm/yr and all the atoms become ionised. Density of Co = 8.83 gm/cc, Atomic weight of Co = 58.93

Answer 1

Solution.

1.

The atomic packing coefficient in the FCC is 74 %, and the atomic packing coefficient in the BCC is 68 %, so the volume of iron will decrease by 6 %.

2.

$0.45 \times m(head) = \rho (Co) \times V(head)$

$V(head) = \frac{4}{3} \times \pi \times R^3$

$0.45 \times \rho(head) = \rho(Co)$

$\rho(head) = 19.62 \ g/cm^3$

$m1(head) = 19.62 \times 1.44 = 28.256 \ g$

$m2(head) = 19.62 \times 1.37 = 26.879 \ g$

$m1(Co) = 0.45 \times 28.256 = 12.72 \ g$

$m2(Co) = 0.45 \times 26.789 = 12.10 \ g$

$N(Co) = \frac{m(Co) \times Na}{Ar(Co)}$

$N1(Co) = 1.30 \times 10^{23}$

$N2(Co) = 1.24 \times 10^{23}$

$\Delta(N) = 6.0 \times 10^{21}$

Answer:

1.

6 %

2.

$\Delta(N) = 6.0 \times 10^{21}$