Problem 1. Aluminum and tantalum have close atomic radii, 0.1431 nm and 0.1430 n

Question

Problem 1. Aluminum and tantalum have close atomic radii, 0.1431 nm and 0.1430 nm, respectively. Tantalum has a BCC crystal structure, and aluminum has a FCC crystal structure.

(a) Which of these metals has the higher atomic packing factor?

(b) Which of these metals has the higher theoretical density?

Problem 2. A common aluminum alloy, Al-2024, contains 4.4 wt.% copper, 1.5 wt. % magnesium, and 0.6 wt. % manganese. The balance is aluminum. What is the atomic % of the various elements in this alloy?

Explanation / Answer

1.a) Tanatalum has a BCC structure and hence has an atomic packing factor (APF) = 0.68

whereas, Al has a FCC structure and hence has an APF = 0.74.

So, Al has a higher APF.

b) theoretical density , rho = (n * A) / (V * Na)

where, n - atoms per unit cell

A - atomic weight

V - volume per unit cell

Na - Avogadro's number = 6.023 * 10^23 atoms/mol

Since, tantalum has a higher atomic number, so it has higher atomic weight and hence a greater theoretical density than Aluminium.

2.Atomic weights -

Cu=64 , Mg = 24 , Mn = 54 , Al = 26

4.4 g Cu = 4.4 / 64 = 0.06875 gram-atom Cu

1.5 g Mg = 1.5 / 24 = 0.0625 gram-atom Mg

0.6 g Mn = 0.6 / 54 = 0.0111 gram-atom Mn

93.5 g Al = 93.5 / 26 = 3.596 gram-atom Al

total = 3.73835 gram-atom

So,

Cu = 0.06875 / 3.73835 = 1.84 atomic %

Mg = 0.0625 / 3.73835 = 1.67 atomic %

Mn = 0.0111/ 3.73835 = 0.297 atomic %

Al = 3.596 / 3.73835 = 96.19 atomic %

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