The energy released when $\frac{7}{17.13} \mathrm{~kg}$ of ${}_{3}^{7} \mathrm{Li}$ is converted into ${}_{2}^{4} \mathrm{He}$ by proton bombardment is $\alpha \times 10^{32} \mathrm{eV}$. The value of $\alpha$ is $\_\_\_\_$. (Nearest integer) (Mass of ${}_{3}^{7} \mathrm{Li}=7.0183 \mathrm{u}$, mass of ${}_{2}^{4} \mathrm{He}=4.004 \mathrm{u}$, mass of proton $=1.008 \mathrm{u}$ and $1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2}$ and Avogadro number $=6.0 \times 10^{23}$)

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Modern Physics

The energy released when

\frac{7}{17.13} kg

_{3}^{7} Li

is converted into

_{2}^{4} He

by proton bombardment is

α \times 1 0^{32} eV

. The value of

α

____

. (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 1 0^{23}

)

Answer: 6

Solution

The reaction

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

has mass defect

Δ m = (7.0183 + 1.008) - 2 (4.004) = 0.0183

u, releasing

0.0183 \times 931 = 17.0

MeV per reaction. The number of Li nuclei in

\frac{7}{17.13}

kg is

\frac{7000/17.13}{7} \times 6.0 \times 1 0^{23} = \frac{6.0 \times 1 0 ^{26}}{17.13} = 3.50 \times 1 0^{25}

. The total energy is

17.0 \times 1 0^{6} \times 3.50 \times 1 0^{25} \approx 6 \times 1 0^{32}

eV, so

α = 6

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Integer answer

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