Question

In the fission of a part of uranium there is loss in mass of 0.5g. How many kilo-watt hour energy will be obtained ?

Answer 1

Answer on Question #63137, Physics / Quantum Mechanics

Question:

In the fission of a part of uranium there is loss in mass of 0.5g. How many kilo-watt hour energy will be obtained?

Solution:

Let's consider one of the possible ways of uranium's fission:

$$U^{235} + {}_0n^1 \rightarrow Y^{89} + Nd^{144} + 3 \cdot {}_0n^1$$

At first we calculate the energy released in the fission of one nucleus. Einstein's formula for mass–energy equivalence looks like this: $E = mc^2$. For the left part of the reaction we obtain

$$E_{in} = (235.117 + 1.00866) \cdot 931.5 \, \text{MeV} = 219951 \, \text{MeV}.$$

For the right part of the reaction:

$$E_{out} = (88.937 + 143.956 + 3 \cdot 1.00866) \cdot 931.5 \, \text{MeV} = 219758.5 \, \text{MeV}.$$

So, for one nucleus $E_0 = E_{out} - E_{in} = 192.5 \, \text{MeV} = 192.5 \cdot 10^6 \, \text{eV} = 192.5 \cdot 10^6 \cdot 1.6 \cdot 10^{-19} \, \text{J} = 3.08 \cdot 10^{-11} \, \text{J}$.

Now we must calculate the number $(N)$ of nuclei in $0.5\,\text{g}$ of uranium.

$$N = \frac{m}{m_{\text{nucleus}}} \cong \frac{m}{u \cdot A}$$

where

- m is the loss of mass (0.5g or $5 \cdot 10^{-4} \, \text{kg}$),

- u — unified atomic mass unit ($1.66 \cdot 10^{-27} \, \text{kg}$),

- A — total number of nucleons (235).

Then $N \cong \frac{m}{u \cdot A} = \frac{5 \cdot 10^{-4}}{1.66 \cdot 10^{-27} \cdot 235} = 1.28 \cdot 10^{21}$.

And total energy: $E = E_0 \cdot N = 3.08 \cdot 10^{-11} \, \text{J} \cdot 1.28 \cdot 10^{21} = 3.94 \cdot 10^{10} \, \text{J}$

One kilo-watt hour $1 \, \text{kWh} = 1000 \, \text{W} \cdot 3600 \, \text{s} = 3.6 \cdot 10^6 \, \text{J}$

So $E \, (\text{kWh}) = \frac{3.94 \cdot 10^{10} \, \text{J}}{3.6 \cdot 10^6 \, \text{J}} = 1.09 \cdot 10^4 \, \text{kWh}$

Answer:

$$1.09 \cdot 10^4 \, \text{kWh}$$

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