Question

In a reactor, 2 kg of ₉₂U²³⁵ fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N = $6.023\times {10}^{26}$per kilo mole and $1eV=1.6\times {10}^{-19}J$. close to

• A. a) 60 MW
• B. b) 54 MW
• C. c) 125 MW
• D. d) 35 MW

Answer 1

The correct option is A

a) 60 MW

Step 1: Given terms:

Avogadro number, $N_A$ = $6.023\times {10}^{26}perkilomole$

$1eV=1.6\times {10}^{–19}J$

$$\begin{gathered} Massnumberofuranium,m=235gm \\ Totalmass.M=2kg \end{gathered}$$

Step 2: Formula used

$n\left(moles\right)=\frac{M}{m}$

$Numberofnucleus\left(N\right)=N_A\times n$

$Power=\frac{N}{t\left(sec\right)}$

Step 3: Find the number of moles

The number moles can be found out by using the formula:

$n\left(moles\right)=\frac{M}{m}$

Where M is the given mass and m is the molar mass

$n\left(moles\right)=\frac{2kg}{235gm}=\frac{2000}{235}$

Step 4: Find the number of nucleus

By substituting the given values in the formula to find number of nucleus, we get:

$$\begin{gathered} Numberofnucleus=6.022\times {10}^{23}\times \frac{2000}{235} \\ =51.25\times {10}^{23} \\ =102.5\times {10}^{25}MeV \end{gathered}$$

Now, to covert eV into Joules:

$$\begin{gathered} \mathrm{N}=102.5\times {10}^{25}\times {10}^6\times 1.6\times {10}^{-16}J \\ =164\times {10}^{6}MJ \end{gathered}$$

Step 5: Find the required value of power

By substituting the given values in the formula to find power, we get:

$Power=\frac{N}{t\left(sec\right)}$

$$\begin{gathered} Power=\frac{164\times {10}^{6}MJ}{30\times 24\times 60\times 60sec} \\ =0.063\times {10}^{3}MW \\ =60MW \end{gathered}$$

Hence the answer is option a ) 60 MW