Concept explainers
Interpretation:
The power output generated at different times due to the decay of
Concept introduction:
The amount of a particular radioactive isotope left after time t is given as:
Here,
is the rate constant for the radioactive decay,
Nearly all radioactive decays are of first order and the rate constant is given as:
Here,
is the half-life of the radioactive substance.
The rate of decay for a particular isotope is given by the rate law as:
Here,
is the rate constant for the radioactive decay and

Answer to Problem 3SEPP
Correct answer: Option (d).
Explanation of Solution
Given information: Initial mass of
Energy of the alpha particle per decay
J
At first, calculate the number of
of sample.
It is known that one mol of any substance is equivalent to its molar mass.
Thus, the conversion factor for
Moreover, it is also known that one mol of any substance is equivalent to Avogadro’s number.
Thus, the conversion factor for
Also, one gram is equivalent to
Thus, the conversion factor is
Hence, by using the conversion factor, the number of atoms of
present in
Thus, the initial number of atoms in one milligram of
is
Now, from the result of SEPP question 2, it is clear that one atom of
decays to give one alpha particle.
Hence,
of
decay to give
Now, the activity of
Substitute the value of rate constant from the result of SEPP question 1 to get:
The energy of the alpha particles is as follows:
It is known that one watt is equivalent to
Thus, the conversion factor is
The conversion of energy is as follows:
Thus, the power dissipated at
Similarly, the power dissipated at
At first, convert the given time into seconds as follows:
The relation between days and year is represented below.
Thus, the conversion factor is
Similarly,
Thus, the conversion factor is
Also,
Thus, the conversion factor is
Now, convert the given time in seconds as follows:
The number of atoms of
decaying in ten years can be calculated as follows:
Here,
Substitute the values of
The activity of
Substitute
The energy of the alpha particles is calculated as follows:
It is known that one watt is equivalent to
milliwatts.
Thus, the conversion factor is
Energy is converted as follows:
Thus, the power dissipated at
Hence, option (d) is correct.
Reasons for the incorrect option:
Option (a) is incorrect because the power dissipated at
Option (b) is incorrect because the power dissipated at
Option (c) is incorrect because the power dissipated at
Hence, options (a), (b), and (c) are incorrect.
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Chapter 20 Solutions
BURDGE CHEMISTRY VALUE ED (LL)
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