AP Physics 2 Exam Question

Consider a nuclear reaction in which a neutron collides with a uranium-235 nucleus resulting in the formation of a krypton-92 nucleus, three neutrons, and the release of energy. The nuclear reaction can be represented as:

$\displaystyle\mathrm{n}+{}^{235}\mathrm{U}\rightarrow{}^{92}\mathrm{Kr}+3\mathrm{n}+\mathrm{energy}$

Given that the rest masses of the particles involved are:

$\begin{align*} \mathrm{mass}({}^{92}\mathrm{Kr})& =91.531 ,u \ \mathrm{mass}(\mathrm{n})& =1.0087 ,u \ \mathrm{mass}({}^{235}\mathrm{U})& = 235.0439 ,u \end{align*}$

where $\mathrm{u}$ denotes atomic mass unit.

(a) Calculate the total mass of the products (excluding energy).

(b) Determine the energy released in this nuclear reaction.

(c) Show that the energy released can be related to the mass defect using Einstein's mass-energy equivalence equation, $\displaystyle\Delta E=\Delta m\cdot c^{2}$, where $\displaystyle c=3.00\times 10^{8}\ \mathrm{m/s}$ is the speed of light in vacuum.

Answer and Step-by-Step Explanation

(a) To calculate the total mass of the products, we need to sum the masses of the individual particles.

Mass of krypton-92 nucleus:

$\begin{align*} \mathrm{mass}({}^{92}\mathrm{Kr})& =91.531 ,u \end{align*}$

Mass of 3 neutrons:

$\begin{align*} \mathrm{mass}(\mathrm{n})& =1.0087 ,u \end{align*}$

Total mass of the products:

$\begin{align*} \mathrm{Total\ mass}(\text{excluding energy})& = \mathrm{mass}({}^{92}\mathrm{Kr}) + 3 \cdot \mathrm{mass}(\mathrm{n}) \ & = 91.531,u + 3 \cdot 1.0087,u \ & = 91.531,u + 3.0261,u \ & = 94.5571,u \end{align*}$

Therefore, the total mass of the products (excluding energy) is $\boxed{94.5571\,u}$.

(b) To determine the energy released in this nuclear reaction, we can use Einstein's mass-energy equivalence equation, $\displaystyle\Delta E=\Delta m\cdot c^{2}$, where $\displaystyle c=3.00\times 10^{8}\ \mathrm{m/s}$ is the speed of light in vacuum.

In this nuclear reaction, the mass of the reactants is $\mathrm{mass}({}^{235}\mathrm{U})=235.0439\,u$.

The mass of the products (excluding energy) is given as $\mathrm{Total\ mass}(\text{excluding energy})=94.5571\,u$.

To find the mass difference, we subtract the mass of the products from the mass of the reactants:

$\begin{align*} \Delta m& = \mathrm{mass}({}^{235}\mathrm{U}) - \mathrm{Total\ mass}(\text{excluding energy}) \ & = 235.0439,u - 94.5571,u \ & = 140.4868,u \end{align*}$

Now, using Einstein's equation $\displaystyle\Delta E=\Delta m\cdot c^{2}$, we can calculate the energy released:

$\begin{align*} \Delta E& = \Delta m \cdot c^{2} \ & = 140.4868,u\cdot (3.00\times 10^{8}\ \mathrm{m/s})^2 \ & = 1.26338\times 10^{24}\ \mathrm{J} \end{align*}$

Therefore, the energy released in this nuclear reaction is $\boxed{1.26338\times 10^{24}\ \mathrm{J}}$.

(c) In part (b), we have already calculated the energy released in the nuclear reaction. Now, we will show that this energy can be related to the mass defect using Einstein's mass-energy equivalence equation, $\displaystyle\Delta E=\Delta m\cdot c^{2}$.

Using the same values as in part (b), we have:

$\begin{align*} \Delta m& = 140.4868,u \ c& = 3.00\times 10^{8}\ \mathrm{m/s} \end{align*}$

Substituting these values into the equation $\displaystyle\Delta E=\Delta m\cdot c^{2}$:

$\begin{align*} \Delta E& = \Delta m \cdot c^{2} \ & = 140.4868,u\cdot (3.00\times 10^{8}\ \mathrm{m/s})^2 \ & = 1.26338\times 10^{24}\ \mathrm{J} \end{align*}$

This confirms that the energy released in the nuclear reaction, calculated using Einstein's equation, is the same as the energy calculated in part (b).

Hence, we have shown that the energy released can be related to the mass defect using Einstein's mass-energy equivalence equation.