Chapter 32, Problem 10TP

To determine

(a)

To Find:

The values of x and y

Answer to Problem 10TP

From the perfect fission reaction of uranium −

• The number of protons in krypton $\text{y = 36}$
• The reaaction realeses extra neutrons $\text{x = 3}$

Explanation of Solution

Given:

The reaction of nuclear fission is

  $\text{n}+\text{U}\to \text{U}\to \text{B}\text{a +}\text{K}\text{r + x }\text{n}$

Formula Used:

An unstable atom is splited into two or more picses of stable atomes and it also realeses the high amount of energy. Also there is release of an extra neutrons.

When one neutron $\left(\text{n}\right)$ is gets added to unstable atoms $\left(\text{U}\right)$, it forms super stable atom $\left(\text{U}\right)$. It can splited into two more stable atoms as $\left(\text{B}\text{a }\right)$ and $\left(\text{K}\text{r}\right)$ with the extra neutron $\left(\text{x }\text{n}\right)$.

The nuclear fiision reaction of Uranium is −

  $\text{n}+\text{U}\to \text{U}\to \text{B}\text{a +}\text{K}\text{r + x }\text{n}$

Here,

Number of Neutrons in Uranium − $\text{236}$

Number of protons in Uranium − $\text{92}$

Number of Neutrons in Barium− $141$

Number of protons in Barium − $56$

Number of Neutrons in Krypton− $92$

Number of protons in Krypton − $\text{y}$

Extra number of neutron - $\text{x}$

Calculation:

From the nuclear reaction-

  $\text{n}+\text{U}\to \text{U}\to \text{B}\text{a +}\text{K}\text{r + x }\text{n}$

We can use the following realtion to determine the extra realaese number of neutrons as $\text{x}$,

  $\begin{array}{l}\text{236 =141+ 92 + x}\left(\text{1}\right)\\ \text{x = 3}\end{array}$

And to determine number of protons in krypton, we can use -

  $\begin{array}{l}\text{92 = 56+ y + x}\left(\text{0}\right)\\ \text{y = 36}\end{array}$

Conclusion:

Hence, the reaction realaese the extra $3$ neutrons. And number of protons in krypton is $36$.

Put the calculated values and the fission reaction of Uranium is as −

  $\text{n}+\text{U}\to \text{U}\to \text{B}\text{a +}\text{K}\text{r + 3}\text{n}$

To determine

(b)

To calculate:

The correct unit of energy released in fission nuclear reaction found by students and conversion of it into the $\text{MeV}$.

Answer to Problem 10TP

From the equation of mass,the energy released equation is found out by

  ${\text{E=mc}}^{\text{2}}$

And after that, it is converted into the proper unit which is

  $\text{E=2}{\text{.689×10}}^{\text{-8}}\text{mJ}$

So, the right unit forcalculated energy is $\text{mJ}$.

After that, it has to transform into a given unit of $\text{MeV}$ by the help of formula−

  $\begin{array}{l}\text{E=}\left(\text{2}{\text{.689×10}}^{\text{-11}}\right)\left(\frac{\text{1eV}}{\text{1}{\text{.6×10}}^{\text{-19}}\text{J}}\right)\left(\frac{{\text{1×10}}^{\text{-6MeV}}}{\text{1eV}}\right)\\ \text{ =168MeV}\end{array}$

Explanation of Solution

Given:

Mass of $\text{U}\text{ = 235}\text{.04 u}$,

Mass of $\text{B}\text{a = 140}\text{.91 u}$,

Mass of $\text{K}\text{r = 91}\text{.93 u}$,

Mass of $\text{n = 1}\text{.01 u}$,

Value of energy released - $\text{E= 2}{\text{.689×10}}^{\text{-8}}$.

Formula used:

1. The total mass of parent and neutron mass -

  ${\text{m}}_{\text{parent}}{\text{=m}}_{\text{Uranium}}{\text{+m}}_{\text{Neutron}}$

2. The products that have total mass,

  ${\text{m}}_{\text{Products}}{\text{=m}}_{\text{Ba}}{\text{+m}}_{\text{Kr}}{\text{+m}}_{\text{neutron}}$

3. The difference of mass of parent nuclei and products (mass lost) −

  ${\text{Δm=m}}_{\text{Parent}}{\text{-m}}_{\text{Product}}$

4. The mass lost unit from $\text{u}$ to $\text{Kg}$-

  $\text{Δm=}\left(\text{0}\text{.18u}\right)\left(\frac{\text{1}{\text{.66×10}}^{\text{-27}}\text{Kg}}{\text{1u}}\right)$

5. Energy released,

  ${\text{E=Δmc}}^{\text{2}}$

6. For converting,

  $\begin{array}{l}\text{E=}\left(\text{2}{\text{.689×10}}^{\text{-11}}\right)\left(\frac{\text{1eV}}{\text{1}{\text{.6×10}}^{\text{-19}}\text{J}}\right)\left(\frac{{\text{1×10}}^{\text{-6MeV}}}{\text{1eV}}\right)\\ \end{array}$

Calculation:

The total mass of parent and neutron mass -

  ${\text{m}}_{\text{parent}}{\text{=m}}_{\text{Uranium}}{\text{+m}}_{\text{Neutron}}$

Put the given values in the equation

  ${\text{m}}_{\text{parent}}\text{=235}\text{.04 u+1}\text{.01 u }$

  ${\text{m}}_{\text{parent}}\text{=236}\text{.05 u}$

Total mass products,

  ${\text{m}}_{\text{Products}}{\text{=m}}_{\text{Ba}}{\text{+m}}_{\text{Kr}}{\text{+m}}_{\text{neutron}}$

Put the values

Mass of $\text{B}\text{a = 140}\text{.91 u}$,

Mass of $\text{K}\text{r = 91}\text{.93 u}$,

Mass of $\text{n = 1}\text{.01 u}$,

  ${\text{m}}_{\text{Products}}\text{=140}\text{.91u+91}\text{.93u+3}\left(\text{1}\text{.01u}\right)$

  $\text{=235}\text{.87 u}$

The difference of mass of parent nuclei and products (mass lost) −

  ${\text{Δm=m}}_{\text{Parent}}{\text{-m}}_{\text{Product}}$

Put the values of

  ${\text{m}}_{\text{Parent}}\text{=235}\text{.87u }$

  ${\text{m}}_{\text{product}}\text{= 235}\text{.87u}$

  $\text{Δm=236}\text{.05u-235}\text{.87 u }$

  $\text{=0}\text{.18u}$

The mass lost unit from $\text{u}$ to $\text{Kg}$ -

  $\text{Δm=}\left(\text{0}\text{.18u}\right)\left(\frac{\text{1}{\text{.66×10}}^{\text{-27}}\text{Kg}}{\text{1u}}\right)$

  $\text{=0}{\text{.2988×10}}^{\text{-27}}\text{Kg}$

Energy released,

  ${\text{E=Δmc}}^{\text{2}}$

put the values of,

  $\text{Δm= 0}{\text{.2988×10}}^{\text{-27}}\text{Kg}$,

  ${\text{c= 3×10}}^{\text{8}}\text{m/s}$

  ${\text{E=Δmc}}^{\text{2}}$

  $\begin{array}{l}\text{E=}\left(\text{0}{\text{.2988×10}}^{\text{-27}}\text{J}\right){\left({\text{3×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}\\ \end{array}$

  $\begin{array}{l}\text{E=}\left(\text{2}{\text{.689×10}}^{\text{-11}}\text{J}\right)\left(\frac{{\text{1×10}}^{\text{3}}\text{mJ}}{\text{1J}}\right)\\ \end{array}$

  $\text{=2}{\text{.689×10}}^{\text{-11}}\text{mJ }$

Thus the correct unit for energy released $\text{=2}{\text{.689×10}}^{\text{-11}}\text{mJ }$

For changing unit from $\text{mJ }$ to $\text{MeV}$,

  $\begin{array}{l}\text{E=}\left(\text{2}{\text{.689×10}}^{\text{-11}}\right)\left(\frac{\text{1eV}}{\text{1}{\text{.6×10}}^{\text{-19}}\text{J}}\right)\left(\frac{{\text{1×10}}^{\text{-6MeV}}}{\text{1eV}}\right)\\ \text{E=168MeV}\\ \end{array}$

Conclusion:

Thus the total energy released from all the data is calculated and it is then transformed into another unit which is $\text{168MeV}$.