Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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 hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.