Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

At 1115 degrees Celcius, where iron is still a solid (melting point 1538 degrees Celcius), the unit cell for the most stable crystal lattice of the metal is face- centered cubic (fcc) with an edge length of 362 pm. What is the atomic radius of iron at this temperature?

Sagot :

Answer: The atomic radius of iron is 128 pm.

Explanation:

To calculate the radius of the metal having FCC crystal lattice, the relationship between edge length and radius follows:

[tex]4r=\sqrt{2}a[/tex]

Where,

a = edge length = 362 pm

r = atomic radius of iron = ?

Plugging values in above equation, we get:

[tex]r=\frac{\sqrt{2}\times 362}{4}\\\\a=128pm[/tex]

Hence, the atomic radius of iron is 128 pm.

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.