91Ó°ÊÓ

The mass of a molecule equal to the sum of the masses of all the atoms included in the molecule is known as molecular mass.

a)

Let us solve the given problem.

To approach this problem, we must first know what the Avogadro number is and what the definition of the molar mass is. We remember that molar mass is the mass per mole ratio. We can write the following relation for the mass:

\(m = \left( {{\rm{molar mass}}} \right)\left( {{\rm{amount of substance}}} \right) = Mn\)

The Avogadro number, which has a unit of one reciprocal mol, denotes the number of particles contained in one mol of mass. It’s value is

\({N_{\rm{A}}} = 6.02214076 \times {10^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)

So to relate the number of particles with mol, we have Avogadro number as a proportionality factor:

\(N = n{N_A}\)

Or expressing it with the first relation we wrote:

\(N = \frac{m}{M}{N_A}\)

Putting in all of the numbers from the problem statement we have:

\(N = \left( {\frac{{1\;\,{\rm{kg}}}}{{50 \times {{10}^{ - 3}}\;{\rm{kg/mol}}}}} \right)\left( {6.022 \times {{10}^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}} \right)\)

And, we have the result of:

\(N = 1.204 \times {10^{25}}\)

Therefore, \(1.204 \times {10^{25}}\) molecules are there in \(1.00\,Kg\) of food.

b)

Consider the given problem and solve.

We can first convert Sievert to rad with\(1\;\,{\rm{Sv}} = 100\,{\rm{rad}}\)and we know how much rad represents energy per kilogram\(1\,{\rm{rad}} = 0.0100\;{\rm{J/kg}}\).

So following this we can see that \(1000{\rm{ }}Sv\)has\(100000{\rm{ }}rad\).So putting that into energy per kilogram we got\(\left( {100000 \times 0.01\;\,{\rm{J/kg}}} \right)\)which gives a\(1000\;{\rm{J/kg}}\). Since energy is simply in this case radiation dose times our goal mass:

\(E = {\rm{ dose }} \times m\)

To get this in energy we got simply to convert this to\({\rm{eV}}\). And we know that\(1\,{\rm{eV}} = 1.60 \times {10^{ - 19}}\;{\rm{J}}\):

\(E = \left( {1000\;{\rm{J}}} \right) \times \left( {1.60 \times {{10}^{ - 19}}\;{\rm{J}}} \right)\)

Which gives the result of:

\(E = 6.25 \times {10^{21}}\;{\rm{J}}\)

To get how many ion pair is created we know that the energy to create one ion pair is\(\Delta E = 32.0\,{\rm{eV}}\). So the number of created pair will be the ratio of total energy relative to the energy required to create a one pair:

\(n = \frac{E}{{\Delta E}}\)

Putting in the numbers we have:

\(n = \frac{{6.25 \times {{10}^{21}}\,{\rm{eV}}}}{{32.0\,{\rm{eV}}}}\)

Hence, gives the result of:

\(n = 1.95 \times {10^{20}}\).

c)

Calculating the ratio of ion-pairs

We find the ratio of ion-pairs relative to the molecules. In part a, we have calculated have many molecules are here, which is\(N = 1.204 \times {10^{25}}\), so we just put in the numbers:

\(\begin{aligned} {\rm{Ratio }} &= \frac{{1.95.0 \times {{10}^{20}}}}{{1.204 \times {{10}^{25}}}}\\{\rm{Ratio }} &= 1.62 \times {10^{ - 5}}\end{aligned}\)

Therefore, the ratio is \(1.62 \times {10^{ - 5}}\).

d)

We want to know how many parts per billion in a new 2000 compound, if the new compound is the result of the pair recombination. We just divide the number of the ion pair by 2000 to get the number and expression is in the parts per billion form:

\(\frac{{1.62 \times {{10}^{ - 5}}}}{{2000}} = 8.098 \times {10^{ - 9}}\).

Hence, it gives \(8.098\) parts per billion.