A metal of atomic mass 108 crystallises in an fcc lattice and has density 10.4 g cm^-3. Using N_A=6.022x10^23 mol^-1 and sqrt2=1.414, find the atomic radius in pm (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss071

A metal of atomic mass $108$ crystallises in an fcc lattice and has density $10.4 g c m^{- 3}$ . Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ and $2 = 1.414$ , find the atomic radius in pm (nearest integer).

Numerical answer type, answer in pm, tolerance 1.

a^{3} = \frac{4 \times 108}{6.022 \times 1 0 ^{23} \times 10.4} = 6.90 \times 1 0^{- 23} c m^{3}

, so

a = 410.1 pm

. In fcc

r = a / (22) = 410.1/ (2 \times 1.414) = 145.0 \approx 145 pm