Question

aluminum crystallizes in a face - centered cubic (fcc) structure with a lattice parameter a = 0.405 nm.
(1) calculate the linear density of aluminum atoms along the 110 direction --> ld110
(2) compare this to the linear density along the 100 direction --> ld100
(3) compare these values to explain (in one sentence) why slip occurs in the close - packed direction 110. note: \slip\ refers to the way metals and other crystalline materials plastically deform under stress.
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Explanation:

Step1: Calculate linear density for [110] direction

For FCC structure, in [110] direction, the length of the vector $l_{[110]}=\sqrt{2}a$. The number of atoms $n_{[110]} = 2$. Linear density $LD_{[110]}=\frac{n_{[110]}}{l_{[110]}}=\frac{2}{\sqrt{2}a}$. Substituting $a = 0.405$ nm, we have $LD_{[110]}=\frac{2}{\sqrt{2}\times0.405\times10^{- 9}}\text{ atoms/m}\approx3.47\times10^{9}\text{ atoms/m}$.

Step2: Calculate linear density for [100] direction

In [100] direction, the length of the vector $l_{[100]}=a$. The number of atoms $n_{[100]} = 1$. Linear density $LD_{[100]}=\frac{n_{[100]}}{l_{[100]}}=\frac{1}{a}$. Substituting $a = 0.405$ nm, we get $LD_{[100]}=\frac{1}{0.405\times10^{-9}}\text{ atoms/m}\approx2.47\times10^{9}\text{ atoms/m}$.

Step3: Compare and explain slip

Since $LD_{[110]}>LD_{[100]}$, there are more atoms per unit - length in the [110] direction, and it is easier for the atoms to slide past each other along the close - packed [110] direction under stress.

Answer:

(1) $LD_{[110]}\approx3.47\times10^{9}\text{ atoms/m}$
(2) $LD_{[110]}>LD_{[100]}$, where $LD_{[100]}\approx2.47\times10^{9}\text{ atoms/m}$
(3) Slip occurs in the [110] direction because it has a higher linear density of atoms compared to the [100] direction, making it easier for atoms to slide past each other under stress.