Question

The ratio of the energy released when \(2.5\times10^{21}\) atoms of uranium undergo nuclear fission and the energy equivalent of \(2\,\mathrm{mg}\) mass of uranium is nearly \[ \text{(Average energy released per fission of uranium nucleus }=200\,\mathrm{MeV}) \]

• A. \(4:5\)
• B. \(8:9\)
• C. \(4:9\)
• D. \(2:3\)

Hint:

Useful relations: \[ \boxed{ E_{\text{fission}}=N\times200\,\mathrm{MeV} } \] and \[ \boxed{ E=mc^2. } \]

Answer 1

The Correct Option is C

Step 1: Calculate the total energy released by fission. Energy released per fission, \[ 200\,\mathrm{MeV} = 200\times10^6\times1.6\times10^{-19} = 3.2\times10^{-11}\,\mathrm{J}. \] Hence, \[ E_1 = 2.5\times10^{21}\times3.2\times10^{-11} = 8\times10^{10}\,\mathrm{J}. \]

Step 2: Calculate the mass-energy equivalent of \(2\,\mathrm{mg}\). \[ m=2\,\mathrm{mg}=2\times10^{-6}\,\mathrm{kg}. \] Using \[ E=mc^2, \] \[ E_2 = 2\times10^{-6}\times(3\times10^8)^2 = 1.8\times10^{11}\,\mathrm{J}. \]

Step 3: Find the required ratio. \[ E_1:E_2 = 8\times10^{10}:1.8\times10^{11} = 8:18 = 4:9. \] Hence, \[ \boxed{4:9} \] Therefore, \[ \boxed{(C)} \] is the correct answer.