Chapter U1.16, Problem 6E

Interpretation Introduction

(a)

Interpretation:

When uranium-235 undergoes nuclear fission then the addition of neutron takes place. The isotope must be identified along with the symbol.

Concept introduction:

In nuclear fission of uranium-235 it is bombarded with fast-moving neutron to generate another isotope of Uranium which further generates lighter nuclides like barium, krypton and 3 neutrons.

Answer to Problem 6E

Uranium-236 is produced when uranium-235 undergoes nuclear fission.

The symbol is ${}_{92}^{236}\text{U}$.

Explanation of Solution

The nuclear reaction in which uranium undergoes nuclear fission with neutron is shown below.

${}_{92}^{235}\text{U}{+}_0^{1}n{\to }_{92}^{236}\text{U}$

Thus, the isotope is uranium-236 with a symbol ${}_{92}^{236}\text{U}$.

Interpretation Introduction

(b)

Interpretation:

The number of protons in the products of nuclear fission of uranium-236 needs to be calculated.

Concept introduction:

In nuclear fission of uranium-235 it is bombarded with fast-moving neutron to generate another isotope of Uranium which further generates lighter nuclides like barium, krypton and 3 neutrons.

Answer to Problem 6E

Total proton is 92.

Explanation of Solution

The nuclear fission of Uranium-235 is represented as follows:

${}_{92}^{235}\text{U}{+}_0^{1}n{\to }_{92}^{236}\text{U}$

Further,

${}_{92}^{236}\text{U}\to \text{B}\text{a}+\text{K}\text{r}+3n$

Here, barium has atomic number 56 and krypton has atomic number 36. Neutron has 0 atomic number. So, the total protons will be:

$\begin{array}{c}p=56+36+0\\ =92\end{array}$

Interpretation Introduction

(c)

Interpretation:

Whether there is any proton change in the nuclear fission of uranium-235 or not must be explained.

Concept introduction:

In nuclear reaction total protons of reactants must be equal to total protons of products.

Answer to Problem 6E

There is no change in protons in the nuclear reaction of uranium-235.

Explanation of Solution

Protons before fission are equal to the sum of protons in uranium-235 and the proton in neutron thus,

$\begin{array}{c}P=92+0\\ =92\end{array}$

Protons after nuclear fission are equal to the sum of protons of barium, protons of krypton and protons in 3 neutrons thus,

$\begin{array}{c}P=56+36+0\\ =92\end{array}$

Thus, there is no change in protons.