Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:


Substance



Atomic Mass (u)



Density (103 Kg m-3)



Carbon (diamond)



12.01



2.22



Gold



197.00



19.32



Nitrogen (liquid)



14.01



1.00



Lithium



6.94



0.53



Fluorine (liquid)



19.00



1.14



[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

Let us suppose that the atoms are ‘tightly packed’ in a solid or liquid phase and


The Density of the substance = D


Atomic mass of the substance = M


Avogadro’s Number, N = 6.022 × 1023


Radius of an atom of the substance = r


Volume of an atom =


Therefore, the volume of the N molecules = ……(1)


Also, the volume of one mole of the substance = …….(2)


Equating the equation (1) and (2) we get :


=



Thus,


(i) Carbon


Given,


Atomic mass of carbon, M = 12.01× 10-3 kg


The Density of the carbon, D = 2.22× 103 kg / m3



r = 1.29× 10-10 m = 1.29 Å


The radius of the carbon atom is 1.29Å.


(ii) Gold


Given,


Atomic mass of Gold atom, M = 197.0× 10-3 kg


The Density of the Gold atom, D = 19.32× 103 kg / m3



r = 1.59× 10-10 m = 1.59 Å


The radius of the Gold atom is 1.59Å.


(iii) Liquid Nitrogen


Given,


Atomic mass of Liquid nitrogen, M = 14.01× 10-3 kg


The Density of the Liquid nitrogen,D = 1.0× 103 kg / m3



r = 1.77× 10-10 m = 1.77 Å


The radius of the Liquid nitrogen atom is 1.77 Å.


(iv) Lithium


Given,


Atomic mass of Lithium, M = 6.94× 10-3 kg


The Density of the Lithium, D = 0.53× 103 kg / m3



r = 1.73× 10-10 m = 1.73 Å


The radius of the Lithium atom is 1.73 Å.


(v) Liquid Fluorine


Given,


Atomic mass of Liquid flourine, M = 19.0× 10-3 kg


The Density of the Liquid flourine,D = 1.14× 103 kg / m3



r = 1.88× 10-10 m = 1.88 Å


The radius of the Liquid flourine atom is 1.88 Å.


The Rough estimates of the size of the atoms thus are:


Substance



Radius (Å)



Carbon (diamond)



1.29



Gold



1.59



Nitrogen (liquid)



1.77



Lithium



1.73



Fluorine (liquid)



1.88



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