Say we got an alloy made of 90 % Au and 10%Cu. The atomic mass of gold is A <m

pevljivuaosyc

pevljivuaosyc

Answered question

2022-05-18

Say we got an alloy made of 90% Au and 10%Cu. The atomic mass of gold is
A A u = 197 g / m o l e
and of copper
A C u = 63.5 g / m o l e
The specific resistivity of gold is
ρ = 22.8 n Ω m
and the Nordheim coefficient is
C = 450 n Ω m
Then by the Nordheim rule we have that the resistivity of this alloy is given as
ρ = ρ A u + C X ( 1 X )
where X=0.1, and represents the percentage of Copper in the alloy. Using the above I got that
ρ = 63.3 n Ω m
, but data shows that the actual resistivity is ρ = 108 n Ω m. What about X? Is it the percentage of atoms in the alloy, or is it the percentage of mass in the alloy?

Answer & Explanation

kazneni3tr2b

kazneni3tr2b

Beginner2022-05-19Added 17 answers

The resistivity is dependent on the number of atoms and so you must find the ratio of copper atoms to the total number of atoms to find X and hence the resistivity of the alloy.
If you do this correctly you should find that the value you have calculated is in agreement with the book value.
The molar fraction of copper (fraction of copper atoms to the copper and gold atoms) is given by
X = ( 10 63.5 ) ( 10 63.5 + 90 197 ) = 394 1537
This comes from the idea that 10 g of copper is 10 63.2 moles of copper which is 10 63.2 × N A atoms of copper where N A is Avagadro's constant.
ρ = 22.8 + 450 × 394 1537 ( 1 394 1537 ) = 108.6

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