Question

choose the most correct answer.

1. Pb has fcc structure with edge length

of unit cell 495 pm. Radius of Pb

atom is


a. 205 pm b. 185 pm

c. 260 pm d. 175 pm​

Answer 1

Answer:(d) 175pm

Explanation: r=a/2√2

But a= 495pm.

Answer 2

Answer:

The radius, r, of the Pb atom measured is $175pm$.

Therefore, option d) $175pm$ is correct.

Explanation:

Given,

The edge length of the unit cell, a = $495pm$

The radius of the Pb atom, r =?

As we know,

• The relation between the radius and the edge length of the fcc ( face-centered cubic ) structure is:
• $\sqrt{2}a = 4r$

After putting the value of 'a' in the equation, we get:

• $\sqrt{2}\times 495 = 4r$
• $4r = 699.93$                ( $\sqrt{2}$ = $1.414$ )
• r = $174.98 pm$

The radius of the Pb atom, r = $174.98pm$ ≈ $175pm$.