A metal of molar mass 100 g mol^-1 crystallises in the fcc lattice and has density 12.0 g cm^-3. Use N_A = 6.022x10^23 mol^-1. Taking sqrt2=1.414, calculate the atomic radius in pm (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss125

A metal of molar mass $100 g mo l^{- 1}$ crystallises in the fcc lattice and has density $12.0 g c m^{- 3}$ . Use $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ . Taking $2 = 1.414$ , calculate the atomic radius in pm (nearest integer).

Numerical answer type, answer in pm, tolerance 1.

a^{3} = \frac{ZM}{ρ N _{A}} = \frac{4 \times 100}{12.0 \times 6.022 \times 1 0 ^{23}} = 5.535 \times 1 0^{- 23} c m^{3}

, so

a = 381.1 pm

. Then

r = \frac{a}{2 2} = \frac{381.1}{2.828} = 134.76 pm \approx 135 pm