An element crystallises in the fcc lattice with unit cell edge 400 pm. Taking sqrt2=1.414, calculate the atomic radius in pm (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3ss119

An element crystallises in the fcc lattice with unit cell edge $400 pm$ . Taking $2 = 1.414$ , calculate the atomic radius in pm (nearest integer).

Numerical answer type, answer in pm, tolerance 1.

Atoms touch along the face diagonal:

4 r = 2 a

, so

r = \frac{a}{2 2} = \frac{400}{2 \times 1.414} = 141.44 pm \approx 141 pm