Question

The strong neutron excess (defined as ) of high-mass nuclei is illustrated by noting that most high-mass nuclides could never fission into two stable nuclei without neutrons being left over. For example, consider the spontaneous fission of a nucleus into two stable daughter nuclei with atomic numbers and . From Appendix F, determine the name of the (a) first and (b) second daughter nucleus. From Fig. 42-5, approximately how many neutrons are in the (c) first and (d) second? (e) Approximately how many neutrons are left over?

Short answer

1. The name of the first daughter nucleus is yttriume .
2. The name of the second daughter nucleus is iodine .
3. There are 50 neutrons in the first daughter nucleus.
4. There are 74 neutrons in the second daughter nucleus.
5. There are 19 neutrons left over after the emission process.

Step-by-step solution

The given data:

Spontaneous emission of the nucleus into two daughter nuclei with atomic numbers 39 and 53 is given.

Understanding the concept of radioactive emission:

As per the problem, the spontaneous emission of the uranium substance into two daughter nuclei cannot be a stable emission process till there are neutrons left over after the decay process. Here, high mass nuclei undergo the process of fission to give two stable daughter nuclei and eject some amount of energy to follow the conservation of energy concept. This conservation law indicates that the internal forces of the nuclei involved in the process are balanced.

Formula:

The number of neutrons of a nucleus is,

N = A - Z ….. (i)

Here, is the atomic mass and Z is the atomic number.

(a) Calculation to identify the first daughter nucleus:

Using the given data and the periodic table (or Appendix F and/or Appendix G), you can see that the atomic number Z = 39 corresponds to the nucleus of the element .

Hence, the first daughter nucleus is yttrium .

(b) Calculation to identify the second daughter nucleus:

Using the given data and the periodic table (or Appendix F and/or Appendix G), you can see that the atomic number Z = 53 corresponds to the nucleus of the element .

Hence, the second daughter nucleus is iodine.

(c) Calculation of the number of neutrons of the first daughter nucleus:

As the molar mass of Yttrium is 88.905 , therefore there will be approximately nucleons in the nucleus of Yttrium. Therefore, the number in Yttrium nucleus is approximately,

89 - 39 = 50

A detailed listing of stable nuclides shows that the stable isotope of yttrium has neutrons (this can also be inferred from the Molar Mass values listed in Appendix F).

Hence, there are 50 neutrons in the first daughter nucleus.

(d) Calculation of the number of neutrons of the second daughter nucleus:

The molar mass of Iodine is 126.9044 g/mol . Therefore there will be approximately 127 nucleons in Iodine nucleus is approximately 127 - 53 = 74 .

A detailed listing of stable nuclides shows that the stable isotope of iodine has 74 neutrons (this can also be inferred from the Molar Mass values listed in Appendix F).

Hence, there are 74 neutrons in the first daughter nucleus.

(e) Calculation of the left over neutrons:

From the given data, the atomic mass of the uranium can be interfered as,

$A_u=235$

Similarly, the atomic mass of yttrium is given using equation (i) and the above data as:

$$\begin{gathered} A_y=\left(39+50\right) \\ =89 \end{gathered}$$

Similarly, the atomic mass of iodine is given using equation (i) and the above data as:

$$\begin{gathered} A_I=\left(53+74\right) \\ =127 \end{gathered}$$

Now, you are given that the fission process is not stable without some neutrons left over. Thus, the number of neutrons is calculated as follows:

$$\begin{gathered} \mathrm{n}=235-89-127 \\ =19 \end{gathered}$$

Hence, the left-over neutrons number is 19 .