Question:

The energy released in the fission of one \(_{92}^{235}{U}\) nucleus is 200 MeV. The energy released in the fission of 235 g of \(_{92}^{235}{U}\) is nearly:

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Remember to convert energy from MeV to Joules using the conversion factor \(1 \, {MeV} = 1.60218 \times 10^{-13} \, {J}\) for accurate calculations.
Updated On: Mar 13, 2025
  • \(15.84 \times 10^{12} \, {J}\)
  • \(19.27 \times 10^{12} \, {J}\)
  • \(13.59 \times 10^{12} \, {J}\)
  • \(17.73 \times 10^{12} \, {J}\)
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The Correct Option is B

Solution and Explanation

Step 1: Calculate the number of \(_{92}^{235}{U}\) nuclei in 235 g. 
One mole of \(_{92}^{235}{U}\) weighs 235 g, and contains Avogadro's number of nuclei, \( N_A = 6.022 \times 10^{23} \). 
Step 2: Convert the energy per nucleus to total energy for all nuclei. Energy per nucleus = 200 MeV
\[ 1 \, {MeV} = 1.60218 \times 10^{-13} \, {J} \] Total energy for all nuclei = \[ 200 \, {MeV/nucleus} \times 6.022 \times 10^{23} \, {nuclei} \times 1.60218 \times 10^{-13} \, {J/MeV} \] Step 3: Perform the calculation. \[ {Total energy} = 200 \times 6.022 \times 10^{23} \times 1.60218 \times 10^{-13} \approx 19.27 \times 10^{12} \, {J} \]

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