The energy released when 7/17.13 kg of 7Li is converted into 4He by proton bombardment is α × 10^32 eV. The value of α is _______. (Nearest integer) (Mass of 7Li = 7.0183 u, mass of 4He = 4.004 u, mass of proton = 1.008 u and 1 u = 931 MeV/c^2 and Avogadro number = 6.0 × 10^23)

The correct answer is Option 6. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q48 – JEE-MAIN 2026

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Home Physics NucleiQuestion #372

The energy released when

\frac{7}{17.13}

kg of

_{3}^{7} Li

is converted into

_{2}^{4} He

by proton bombardment is

α \times 10^{32}

eV. The value of

α

is _______. (Nearest integer) (Mass of

_{3}^{7} Li = 7.0183

u, mass of

_{2}^{4} He = 4.004

u, mass of proton

= 1.008

u and

1 u = 931 {MeV/c}^{2}

and Avogadro number

= 6.0 \times 10^{23}

)

Enter Calculated Numerical Value:

Clue 1 of 3: Nuclear Reaction Energy

The energy released in a nuclear reaction is determined by the mass defect, which is the difference between the total mass of reactants and the total mass of products.

🧠 Full Solution Path

Step 1: Determine the nuclear reaction and mass defect✦ Active

The nuclear reaction described is the bombardment of Lithium-7 with a proton to produce two Helium-4 nuclei:

_{3}^{7} Li +_{1}^{1} p \to 2_{2}^{4} He

The mass defect ( $Δ m$ ) is the difference between the total mass of reactants and the total mass of products:

Δ m = ({mass of}_{3}^{7} Li + mass of proton) - (2 \times {mass of}_{2}^{4} He)

Δ m = (7.0183 u + 1.008 u) - (2 \times 4.004 u)

Δ m = 8.0263 u - 8.008 u = 0.0183 u

Step 2: Calculate the energy released per reaction○ Expand

Using the mass-energy equivalence relation $1 u = 931 {MeV/c}^{2}$ , the energy released per reaction ( $Q$ ) is:

Q = Δ m \times 931 MeV/u = 0.0183 \times 931 MeV

Q = 17.0373 MeV

Convert this energy to electron volts (eV):

Q = 17.0373 \times 10^{6} eV

Step 3: Calculate the total energy released for the given mass○ Expand

The mass of $_{3}^{7} Li$ converted is $\frac{7}{17.13}$ kg. The molar mass of $_{3}^{7} Li$ is $7 g/mol = 7 \times 10^{- 3} kg/mol$ . The number of moles ( $n$ ) of $_{3}^{7} Li$ converted is:

n = \frac{mass}{molar mass} = \frac{7 / 17.13 kg}{7 \times 10^{- 3} kg/mol} = \frac{1000}{17.13} mol

Using Avogadro's number ( $N_{A} = 6.0 \times 10^{23}$ ), the total number of $_{3}^{7} Li$ atoms ( $N$ ) is:

N = n \times N_{A} = \frac{1000}{17.13} \times 6.0 \times 10^{23} atoms

The total energy released ( $E_{t o t a l}$ ) is the product of the number of atoms and the energy released per reaction:

E_{t o t a l} = N \times Q

E_{t o t a l} = (\frac{1000}{17.13} \times 6.0 \times 10^{23}) \times (17.0373 \times 10^{6} eV)

E_{t o t a l} = \frac{6000 \times 17.0373}{17.13} \times 10^{29} eV

E_{t o t a l} \approx 5967.53 \times 10^{29} eV

E_{t o t a l} \approx 5.96753 \times 10^{32} eV

Comparing this with the given form $α \times 10^{32}$ eV, we find $α \approx 5.96753$ . The nearest integer value for $α$ is 6.

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Physics Question 48 – JEE-MAIN 2026

Clue 1 of 3: Nuclear Reaction Energy

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