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Question: Given mass number of gold = 197, \({\text{Density of gold}} = 19.7g/c{m^3}\), \({\text{Avogadro'...

Given mass number of gold = 197,
Density of gold=19.7g/cm3{\text{Density of gold}} = 19.7g/c{m^3},
Avogadro’s number=6×1023{\text{Avogadro's number}} = 6 \times {10^{23}}. The radius of gold atom is approximately:
A) 1.5×10−8m1.5 \times {10^{ - 8}}m
B) 1.7×10−9m1.7 \times {10^{ - 9}}m
C) 1.5×10−10m1.5 \times {10^{ - 10}}m
D) 1.5×10−12m1.5 \times {10^{ - 12}}m

Explanation

Solution

In this question we have the mass of gold and the density of gold atoms and Avogadro’s number. We have to find the radius of the gold atom. As we know that the density is equal to the ratio of mass and volume. First thing that we need to do is to convert the unit of density. We will convert it to the M.K.S. system. In this formula we have to put mass in kilograms, so we have to convert the mass number in kilograms. As we know that the unit of mass number is a.m.u. (Atomic Mass Unit). To convert a.m.u. to kilograms we will use the following relation.
1 a.m.u.=1.66×10−27 kg1{\text{ a}}{\text{.m}}{\text{.u}}{\text{.}} = 1.66 \times {10^{ - 27}}{\text{ }}kg

Complete step by step solution:
Given,
Density of gold=19.7g/cm3{\text{Density of gold}} = 19.7g/c{m^3}
Density of gold=19.7×103kg/m3{\text{Density of gold}} = 19.7 \times {10^3}kg/{m^3}
Given mass number of gold = 197
⇒Mass of gold = Mass number×1.66×10−27\Rightarrow {\text{Mass of gold = Mass number}} \times 1.66 \times {10}^{-27}
⇒Mass of gold =197×1.66×10−27\Rightarrow {\text{Mass of gold =}} 197 \times 1.66 \times {10^{-27}}
⇒Mass of gold = 327.02×10−27kg\Rightarrow {\text{Mass of gold = }}327.02 \times {10^{ - 27}}kg
Let the radius of the gold atom is R.
The density of an atom is given by following formula,
⇒Density = massvolume\Rightarrow {\text{Density = }}\dfrac{{{\text{mass}}}}{{{\text{volume}}}}
The shape of an atom is spherical, so the volume of gold atom will be Volume=43πR3{\text{Volume}} = \dfrac{4}{3}\pi {R^3}
⇒43πR3 = massdensity\Rightarrow \dfrac{4}{3}\pi {R^3}{\text{ = }}\dfrac{{{\text{mass}}}}{{{\text{density}}}}
⇒R3 = mass43π density\Rightarrow {R^3}{\text{ = }}\dfrac{{{\text{mass}}}}{{\dfrac{4}{3}\pi {\text{ density}}}}
⇒R3 = 327.02×10−2743×3.14×19.7×103 \Rightarrow {R^3}{\text{ = }}\dfrac{{327.02 \times {{10}^{ - 27}}}}{{\dfrac{4}{3} \times 3.14 \times 19.7 \times {{10}^3}{\text{ }}}}
⇒R3 = 981.06×10−27247.432×103\Rightarrow {R^3}{\text{ = }}\dfrac{{981.06 \times {{10}^{ - 27}}}}{{247.432 \times {{10}^3}}}
⇒R3 = 3.9649×10−30\Rightarrow {R^3}{\text{ = 3}}{\text{.9649}} \times {\text{1}}{{\text{0}}^{ - 30}}
⇒R = 1.5827×10−10 m\Rightarrow R{\text{ = 1}}{\text{.5827}} \times {\text{1}}{{\text{0}}^{ - 10}}{\text{ m}}

Hence, the radius of gold atom is R = 1.5827×10−10 mR{\text{ = 1}}{\text{.5827}} \times {\text{1}}{{\text{0}}^{ - 10}}{\text{ m}}.

Note: In such a type of question we have to be careful about the units. All the quantities should be in the same system. The relation between a.m.u. and kilograms should be known. The calculation should be done with care. As we know that Gold is a metal and widely used in jewelry. The purest form of gold is 24- Carat. This form of gold is very soft so some other alloy mostly copper is mixed with gold. This makes it firm and durable. This alloy mixed gold is 18 or 19- Caret. It is widely used for jewelry purposes.