
Concept explainers
Interpretation:
The equation for decay and formation of
are to be written. The energy of alpha particle and the energy released from the given amount of
are to be determined.
Concept introduction:
In a balanced radioactive decay,
And,
The energy during the decay of a radioactive substance is given as:
Here,
is the amount of energy released,
is the mass defect which occurs during the course of decay and c is the speed of the light.
The mass defect (
) is given as the difference of total atomic mass of the product and the reactant in the balanced radioactive reaction.

Answer to Problem 102AP
Solution:
(a)
(b)
(c)
(d)
Explanation of Solution
a) The equation for the reaction;
is prepared by bombarding
with neutrons.
It is known that the formation of
Now, apply the balancing rules as follows:
And
On the left hand side of the reaction, sum of atomic number is calculated as follows:
On the right–hand side, the sum of atomic number is
Thus, Atomic number of X is calculated as follows:
Similarly,
On the left–hand side of the reaction, sum of mass number is calculated as follows;
On the right side of the reaction, sum of mass number is
Thus, mass number of X is calculated as follows:
So, X will be
Hence, the balanced reaction is as follows:
b) The equation of decay process;
It is known that the decay of
Now, apply the balancing rules as follows:
And
On the left-hand side of the reaction, sum of atomic number is
On the right-hand side, the sum of atomic number is
Thus, atomic number of X is calculated as follows:
Similarly,
On the left hand side of the reaction, sum of mass number is
On the right side of the reaction, sum of mass number is
Thus, mass number of X is calculated as follows:
So, X will be
Hence, the balanced reaction is as follows:
c) The energy of an emitted
particle in given reaction:
Mass of
is
Mass of
is
Mass of
is
The balanced reaction is given as follows:
Now, the mass defect
is calculated as follows:
Substitute the values of masses in the above expression,
Now, the magnitude of mass defect is taken for further calculation as mass is converted to energy.
Now, it is known that
Thus, the conversion factor is
Hence, the mass defect can be converted to kilogram unit by using the above conversion factor as follows:
The energy during the decay of a radioactive substance is given as follows:
Substitute the values of energy and mass defect in the above expression,
Here, it is assumed that all the energy generated by the mass defect is contributed to the kinetic energy of the alpha particle.
Thus, the kinetic energy of alpha particle is
d) The total energy released by
of
At first, calculate the number of
atoms in
It is known that one micro gram of any substance is equivalent to
Thus, the conversion factor is
It is known that one mol of any substance is equivalent to its molar mass.
Thus, the conversion factor for
is
Similarly, it is known that one mol of any substance is equivalent to Avogadro’s number.
Thus, the conversion factor is
Hence, using the appropriate conversion factor, the number of
atoms in
Now, as per the given information, it is known that half –life of polonium -
is
will decay to form alpha particle. Thus, the number of polonium atoms decayed will be equivalent to the number of alpha particles formed.
Hence, the number of polonium atoms decayed in
Now, use the energy result obtained in part (c) to get the energy of the alpha particle as follows:
Thus, the total energy released from
of polonium is
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Chapter 20 Solutions
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