An element with molar mass $2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}$ forms a cubic unit cell with edge length 405 pm . If its density is $2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, the radius of the element is approximately \_\_\_\_ $\times 10^{-12} \mathrm{~m}$ (to the nearest integer).

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Solid State

An element with molar mass

2.7 \times 1 0^{- 2} kg mol^{- 1}

forms a cubic unit cell with edge length 405 pm . If its density is

2.7 \times 1 0^{3} kg m^{- 3}

, the radius of the element is approximately
____

\times 1 0^{- 12} m

(to the nearest integer).

Answer: 143

From the unit-cell density

d = \frac{Z M}{N _{A} a ^{3}}

, inserting

d = 2.7 \times 1 0^{3}

kg m

^{- 3}

M = 2.7 \times 1 0^{- 2}

kg mol

^{- 1}

and

a = 405

= 4.05 \times 1 0^{- 10}

m gives

Z = 4

, so the metal is face-centred cubic. In an fcc cell the atoms touch along the face diagonal,

4 r = 2 a

, so

r = \frac{2 \times 405}{4} \approx 143

pm, i.e.

143 \times 1 0^{- 12}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Integer answer