The diameter of the zinc atom is $2.6 o A$ . Calculate (a) radius of zinc atom in pm and (b) a number of atoms present in a length of $1.6 cm$ if the zinc atoms are arranged side by side lengthwise.

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Solution

(i)The diameter of zinc atom is $2.6 o A = 2.6 \times 10^{- 10} m$ .
The radius of Zn atom is $\frac{2.6 \times 10^{- 10}}{2} = 1.3 \times 10^{- 10} m = 130 \times 10^{- 12} m = 130 p m$ .
(ii) The number of Zn atoms present on 1.6 cm of length are $\frac{1.6}{2.6 \times 10^{- 8}} = 6.154 \times 10^{7}$ .

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The diameter of the zinc atom is 2.6oA. Calculate (a) radius of zinc atom in pm and (b) a number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

(i)The diameter of zinc atom is 2.6oA=2.6×10−10m.The radius of Zn atom is 2.6×10−102=1.3×10−10m=130×10−12m=130 pm.(ii) The number of Zn atoms present on 1.6 cm of length are 1.62.6×10−8=6.154×107.

The diameter of the zinc atom is $2.6 o A$ . Calculate (a) radius of zinc atom in pm and (b) a number of atoms present in a length of $1.6 cm$ if the zinc atoms are arranged side by side lengthwise.