You are given that Mass of ⁷₃Li = 7.0160u, Mass of ⁴₂He = 4.0026u and Mass of ¹₁H = 1.0079u. When 20 g of ⁷₃Li is converted into ⁴₂He by proton capture, the energy liberated, (in kWh), is: [Mass of nucleon = 1 GeV/c2] (a) 6.82 × 105 (b) 4.5 × 105 (c) 8 × 106 (d) 1.33 × 106

Answer: (d). ⁷₃Li + ¹₁H → 2 ⁴₂He: Δ m = 7.0160 + 1.0079 - 2(4.0026) = 0.0187 u Energy per reaction = 0.0187 × 931.5 ≈ 17.4 MeV Number of Li nuclei in 20 g = 20/7 × 6.02 × 10²³ = 1.72 × 10²⁴ E = 1.72 × 10²⁴ × 17.4 × 1.6 × 10⁻¹³ J ≈ 4.8 × 10¹² J In kWh: 4.8 × 10¹²3.6 × 10⁶ ≈ 1.33 × 10⁶ kWh So the answer is (d) 1.33 × 10⁶ kWh.

You are given that Mass of ⁷₃Li = 7.0160u, Mass of ⁴₂He = 4.0026u and… | JEE Main 2020 (6 Sep, Shift 1)

Q15JEE Main 2020 · 6 SEP · SHIFT 1Difficult

Nuclear Reactions & Q-value

You are given that Mass of ${}^{7}_{3}Li$ = 7.0160u,

Mass of ${}^{4}_{2}He$ = 4.0026u

and Mass of ${}^{1}_{1}H$ = 1.0079u.

When 20 g of ${}^{7}_{3}Li$ is converted into ${}^{4}_{2}He$ by proton capture, the energy liberated, (in kWh), is: [Mass of nucleon = 1 GeV/c2]

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Answer: (d)

${}^{7}_{3}Li + {}^{1}_{1}H \to 2\,{}^{4}_{2}He$ : $\Delta m = 7.0160 + 1.0079 - 2(4.0026) = 0.0187\ \text{u}$
Energy per reaction $= 0.0187 \times 931.5 \approx 17.4\ \text{MeV}$
Number of Li nuclei in 20 g $= \dfrac{20}{7} \times 6.02 \times 10^{23} = 1.72 \times 10^{24}$
$E = 1.72 \times 10^{24} \times 17.4 \times 1.6 \times 10^{-13}\ \text{J} \approx 4.8 \times 10^{12}\ \text{J}$
In kWh: $\dfrac{4.8 \times 10^{12}}{3.6 \times 10^{6}} \approx 1.33 \times 10^{6}\ \text{kWh}$
So the answer is (d) $1.33 \times 10^{6}\ \text{kWh}$ .

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Nuclear Reactions & Q-value question from JEE Main 2020 (6 Sep, Shift 1)

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