24.102 CP - Chemistry: Molecular Nature Of Matter And Change

Question

The scene below depicts a neutron bombarding ${}^{\text{235}}\text{U}$ :

1. Is this an example of fission or of fusion? 

(b) Identify the other nuclide formed. 

(c) What is the most likely mode of decay of the nuclide with Z-55?

Step-by-Step Solution

Answer

1. The process is nuclear fission.
2. The other nuclide formed is  ${}_{\text{92}}{}^{\text{235}}{\text{U + }}_{\text{0}}^{\text{1}}\text{n}{\to }_{\text{55}}^{\text{144}}{\text{Cs + }}_{\text{37}}^{\text{90}}{\text{Rb + 2}}_{\text{0}}^{\text{1}}\text{n}$  .
3. The most likely mode of decay of the nuclide with Z-55  is ${}_{\text{55}}{}^{\text{144}}\text{Cs}$  undergo  decay.

Step 1: Nuclear Fusion

Nuclear fusion is a kind of nuclear reaction in which two lighter nuclei combine to form a heavier nucleus, accompanied with the emission of large amount of energy.

Step 1: Nuclear Fission

Nuclear fission is a nuclear reaction in which a heavier nuclei disintegrates into lighter nuclei along with the liberation of large amount of energy. Nuclear fission proceeds through a chain mechanism. When a unstable nuclei undergo a fission reaction, neutrons are also released. These neutrons formed bombard other unstable nuclei and hence the process continues.

Step 2: Find this is an example of fission or of fusion

(a)

Uranium-235 (a large nucleus) is bombarded with a neutron, splitting into two intermediate-mass nuclei in the process.

Since, a heavier nuclei disintegrates to form to lighter nuclei, the process is nuclear fission.

Step 3: Find the other nuclide formed

Mass and charge are balanced in nuclear reactions.

 ${}_{\text{92}}{}^{\text{235}}{\text{U + }}_{\text{0}}^{\text{1}}\text{n}{\to }_{\text{55}}^{\text{144}}{\text{Cs + }}_{\text{37}}^{\text{90}}{\text{Rb + 2}}_{\text{0}}^{\text{1}}\text{n}$

Thus, the other nuclei formed is ${}_{\text{37}}{}^{\text{90}}\text{Rb}$ . 

Step 4: Find the most likely mode of decay of the nuclide with

Consider the given information:

No. of protons,   

No. of neutrons,  ${\text{(N) in }}_{\text{55}}^{\text{144}}\text{Cs = (144 - 55) = 89}$

Neutron to proton ratio  ${\text{(N/Z) of }}_{\text{55}}^{\text{144}}\text{Cs}$ is  89/55 = 1.6.

The neutron proton ratio (N/Z)  is high and outside the stability band.

Therefore, ${}_{\text{55}}{}^{\text{144}}\text{Cs}$  undergo  decay so that the number of neutrons decreases and the  (N/Z) decreases.