Name: Electrical Properties of Semiconductors (Band Theory)
Uploaded: 2019-07-19
Duration: 8 min 25 s
Description: In semiconductors, there is a small gap between the valence band and the conduction band. Then, how do they conduct electricity? Let’s watch the video and fi...

Question

Calculate the radius of copper atom. The atomic weight of copper is $63.55 g {mol}^{- 1}$ . It crystallises in face centred cubic lattice and has density of $8.93 g {cm}^{- 3}$ at $298 K$ . ( $N_{A} = 6.023 \times 10^{23} {mol}^{- 1}$ ).

Accepted Answer

Given,

Atomic weight of copper is $63.55 g {mol}^{- 1}$ .

The density is $8.93 g {cm}^{- 3}$ .

Avogadro's number is $N_{A} = 6.023 \times 10^{23} {mol}^{- 1}$ .

Type of unit cell is face-centred unit cell. So, $z = 4$ .

$d = \frac{z \times M}{a^{3} \times N_{A}} \Rightarrow a^{3} = \frac{z \times M}{d \times N_{A}} . a^{3} = \frac{4 \times 63.55}{8.93 \times 6.023 \times 10^{23}} = \frac{254.2 \times 10^{- 23}}{53.79} . a^{3} = 47.3 \times 10^{- 24} {cm}^{3} . a = 3.61 \times 10^{- 8} cm = 361 pm .$

The radius can be calculated as,

For FCC, $\sqrt{2} a = 4 r \Rightarrow r = \frac{\sqrt{2} \times a}{4}$ .

$r = \frac{\sqrt{2} \times 361}{4} = \sqrt{2} \times 90.25 . r = 1.414 \times 90.25 = 127.61 pm .$

Calculate the radius of copper atom. The atomic weight of copper is $63.55 g {mol}^{- 1}$ . It crystallises in face centred cubic lattice and has density of $8.93 g {cm}^{- 3}$ at $298 K$ . ( $N_{A} = 6.023 \times 10^{23} {mol}^{- 1}$ ).

Important Questions on The Solid State

Learn and Prepare for any exam you want !

Calculate the radius of copper atom. The atomic weight of copper is 63.55 g mol-1. It crystallises in face centred cubic lattice and has density of 8.93 g cm-3 at 298 K. (NA=6.023×1023 mol-1).

Important Questions on The Solid State

Learn and Prepare for any exam you want !

Calculate the radius of copper atom. The atomic weight of copper is $63.55 g {mol}^{- 1}$ . It crystallises in face centred cubic lattice and has density of $8.93 g {cm}^{- 3}$ at $298 K$ . ( $N_{A} = 6.023 \times 10^{23} {mol}^{- 1}$ ).