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NVAEasyJEE 2023Nuclear Fission & Fusion

JEE Physics 2023 Question with Solution

The energy released per fission of nucleus of 240X^{240}X is 200 MeV200 \, \text{MeV}. The energy released if all the atoms in 120 g120 \, \text{g} of pure 240X^{240}X undergo fission is:

Answer

Correct answer:6

Step-by-step solution

Standard Method

Given: Energy released per fission of nucleus of 240X^{240}X is 200 MeV200 \, \text{MeV} and mass of sample is 120 g120 \, \text{g}.

Find: Total energy released if all atoms undergo fission.

The number of moles of 240X^{240}X in 120 g120 \, \text{g} is

No. of moles=MassMolar Mass=120240=0.5 moles\text{No. of moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{120}{240} = 0.5 \, \text{moles}

The number of nuclei in 0.50.5 moles is

No. of nuclei=0.5⋅NA=0.5⋅6×1023=3×1023\text{No. of nuclei} = 0.5 \cdot N_A = 0.5 \cdot 6 \times 10^{23} = 3 \times 10^{23}

The total energy released is

E=(No. of nuclei)⋅(Energy per fission)E = (\text{No. of nuclei}) \cdot (\text{Energy per fission})

Substitute:

E=(3×1023)⋅(200)=6×1025 MeVE = (3 \times 10^{23}) \cdot (200) = 6 \times 10^{25} \, \text{MeV}

Therefore, the total energy released is 6×1025 MeV6 \times 10^{25} \, \text{MeV}. Hence the numerical answer is 6.

Direct Multiplication

Given: Mass of 240X^{240}X is 120 g120 \, \text{g} and energy per fission is 200 MeV200 \, \text{MeV}.

Find: Total energy released.

Since 120 g120 \, \text{g} of 240X^{240}X corresponds to

120240=12\frac{120}{240} = \frac{1}{2}

mole, the number of nuclei is

12×6×1023=3×1023\frac{1}{2} \times 6 \times 10^{23} = 3 \times 10^{23}

Now multiply by energy released per fission:

3×1023×200=6×1025 MeV3 \times 10^{23} \times 200 = 6 \times 10^{25} \, \text{MeV}

Therefore, the required value is 6.

Common mistakes

  • Using 240240 as the number of nuclei instead of the molar mass is incorrect. 240240 is the mass number and also the molar mass in g mol−1\text{g mol}^{-1} for this calculation. First convert mass into moles using massmolar mass\frac{\text{mass}}{\text{molar mass}}.

  • Forgetting to multiply the number of moles by Avogadro's number gives the wrong count of nuclei. Energy per fission is for one nucleus, so convert moles to nuclei using N=nNAN = nN_A before multiplying by 200 MeV200 \, \text{MeV}.

  • Writing the final answer as 6×10256 \times 10^{25} in the answer box is incorrect for this numerical-value format. The extracted final quantity is 6×1025 MeV6 \times 10^{25} \, \text{MeV}, so the required numerical entry is 6.

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