Medium
The diameter of a zinc atom is $2.6 \, \mathring{A}$. Calculate $(a)$ the radius of the zinc atom in $pm$ and $(b)$ the number of atoms present in a length of $1.6 \, cm$ if the zinc atoms are arranged side by side lengthwise.
$(a)$ Radius of zinc atom $= \frac{\text{Diameter}}{2} = \frac{2.6 \, \mathring{A}}{2} = 1.3 \, \mathring{A} = 1.3 \times 10^{-10} \, m = 130 \, pm$.
$(b)$ Length of the arrangement $= 1.6 \, cm = 1.6 \times 10^{-2} \, m$.
Diameter of zinc atom $= 2.6 \, \mathring{A} = 2.6 \times 10^{-10} \, m$.
Number of zinc atoms $= \frac{\text{Total length}}{\text{Diameter of one atom}} = \frac{1.6 \times 10^{-2} \, m}{2.6 \times 10^{-10} \, m} \approx 6.153 \times 10^7$ atoms.
$(b)$ Length of the arrangement $= 1.6 \, cm = 1.6 \times 10^{-2} \, m$.
Diameter of zinc atom $= 2.6 \, \mathring{A} = 2.6 \times 10^{-10} \, m$.
Number of zinc atoms $= \frac{\text{Total length}}{\text{Diameter of one atom}} = \frac{1.6 \times 10^{-2} \, m}{2.6 \times 10^{-10} \, m} \approx 6.153 \times 10^7$ atoms.
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Similar Questions
Match the following physical quantities with their appropriate units.
Physical Quantity Unit $(A)$ Molarity $(1)$ $g \ mL^{-1}$ $(B)$ Mole fraction $(2)$ $mol$ $(C)$ Mole $(3)$ $Pascal$ $(D)$ Molality $(4)$ $Unitless$ $(E)$ Pressure $(5)$ $mol \ L^{-1}$ $(F)$ Luminous intensity $(6)$ $Candela$ $(G)$ Density $(7)$ $mol \ kg^{-1}$ $(H)$ Mass $(8)$ $N \ m^{-1}$ $(9)$ $kg$
| Physical Quantity | Unit |
|---|---|
| $(A)$ Molarity | $(1)$ $g \ mL^{-1}$ |
| $(B)$ Mole fraction | $(2)$ $mol$ |
| $(C)$ Mole | $(3)$ $Pascal$ |
| $(D)$ Molality | $(4)$ $Unitless$ |
| $(E)$ Pressure | $(5)$ $mol \ L^{-1}$ |
| $(F)$ Luminous intensity | $(6)$ $Candela$ |
| $(G)$ Density | $(7)$ $mol \ kg^{-1}$ |
| $(H)$ Mass | $(8)$ $N \ m^{-1}$ |
| $(9)$ $kg$ |
Medium
View SolutionWhat do you mean by significant figures?
Medium
View SolutionHow many significant figures should be present in the answer of the following calculations?
$(i)$ $\frac{0.02856 \times 298.15 \times 0.112}{0.5785}$
$(ii)$ $5 \times 5.364$
$(iii)$ $0.0125 + 0.7864 + 0.0215$
$(i)$ $\frac{0.02856 \times 298.15 \times 0.112}{0.5785}$
$(ii)$ $5 \times 5.364$
$(iii)$ $0.0125 + 0.7864 + 0.0215$
Medium
View SolutionOne calorie is equal to (in $J$)
Easy
View SolutionObserve the following:
$I) 0.0063$
$II) 132.00$
$III) 1004$
The number of significant figures in $I, II$ and $III$ is respectively:
$I) 0.0063$
$II) 132.00$
$III) 1004$
The number of significant figures in $I, II$ and $III$ is respectively:
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