1. sqrt3/4×288 pm 2. sqrt2/4×288 pm 3. 4/sqrt3×288 pm 4. 4/sqrt2×288 pm

Answer: 1. sqrt3/4×288 pm In a bcc unit cell atoms touch along the body diagonal. Body diagonal =sqrt3 a=4r . r=sqrt3/4a=sqrt3/4×288\ pm≈124.7\ pm Option (2) sqrt2/4 is the fcc relation, so it is wrong here.

Topic: Metallic bonding and hydrogen bonding· Chemical Bonding and Molecular Structure

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is :

1. $\frac{\sqrt3}{4}\times288$ pm

2. $\frac{\sqrt2}{4}\times288$ pm

3. $\frac{4}{\sqrt3}\times288$ pm

4. $\frac{4}{\sqrt2}\times288$ pm

Doubtly Asked 30 Oct 2026 NEET UG 2020 neet Save WhatsApp Report

1 answer

Best answer

Answer: 1. $\frac{\sqrt3}{4}\times288$ pm

In a bcc unit cell atoms touch along the body diagonal. Body diagonal $=\sqrt3\,a=4r$.

$$r=\frac{\sqrt3}{4}a=\frac{\sqrt3}{4}\times288\ \mathrm{pm}\approx124.7\ \mathrm{pm}$$

Option (2) $\sqrt2/4$ is the fcc relation, so it is wrong here.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is :

1 answer

Your answer