Q73 / 75 · JEE Main 2026 (06 April Shift 1) · Physics
The energy released when $\dfrac{7}{17.13}$ kg of $^{7}_{3}\text{Li}$ is converted into $^{4}_{2}\text{He}$ by proton bombardment is $\alpha \times 10^{32}$ eV. The value of $\alpha$ is _______. (Nearest integer) (Mass of $^{7}_{3}\text{Li} = 7.0183$ u, mass of $^{4}_{2}\text{He} = 4.004$ u, mass of proton $= 1.008$ u and $1$ u $= 931$ MeV/c$^2$ and Avogadro number $= 6.0 \times 10^{23}$)
Correct answer: 6
Solution
The nuclear reaction for the proton bombardment of Lithium is:
$^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow 2 ^{4}_{2}\text{He}$
The mass defect ($\Delta m$) for the reaction is:
$\Delta m = m(^{7}_{3}\text{Li}) + m(^{1}_{1}\text{H}) - 2m(^{4}_{2}\text{He})$
$\Delta m = 7.0183 + 1.008 - 2(4.004)$
$\Delta m = 8.0263 - 8.008 = 0.0183 \text{ u}$
Energy released per reaction ($Q$) is:
$Q = \Delta m \times 931 \text{ MeV} = 0.0183 \times 931 \text{ MeV} = 17.0373 \text{ MeV}$
Number of moles of $^{7}_{3}\text{Li}$ in $\dfrac{7}{17.13} \text{ kg}$:
$n = \dfrac{\dfrac{7}{17.13} \times 10^3 \text{ g}}{7 \text{ g/mol}} = \dfrac{1000}{17.13} \text{ mol}$
Number of atoms of $^{7}_{3}\text{Li}$:
$N = n \times N_A = \dfrac{1000}{17.13} \times 6.0 \times 10^{23} = \dfrac{6}{17.13} \times 10^{26}$
Total energy released ($E$) is:
$E = N \times Q = \left(\dfrac{6}{17.13} \times 10^{26}\right) \times (17.0373 \times 10^6 \text{ eV})$
$E = \dfrac{102.2238}{17.13} \times 10^{32} \text{ eV} \approx 5.967 \times 10^{32} \text{ eV}$
Rounding to the nearest integer, we get $\alpha = 6$.
Answer: $6$
$^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow 2 ^{4}_{2}\text{He}$
The mass defect ($\Delta m$) for the reaction is:
$\Delta m = m(^{7}_{3}\text{Li}) + m(^{1}_{1}\text{H}) - 2m(^{4}_{2}\text{He})$
$\Delta m = 7.0183 + 1.008 - 2(4.004)$
$\Delta m = 8.0263 - 8.008 = 0.0183 \text{ u}$
Energy released per reaction ($Q$) is:
$Q = \Delta m \times 931 \text{ MeV} = 0.0183 \times 931 \text{ MeV} = 17.0373 \text{ MeV}$
Number of moles of $^{7}_{3}\text{Li}$ in $\dfrac{7}{17.13} \text{ kg}$:
$n = \dfrac{\dfrac{7}{17.13} \times 10^3 \text{ g}}{7 \text{ g/mol}} = \dfrac{1000}{17.13} \text{ mol}$
Number of atoms of $^{7}_{3}\text{Li}$:
$N = n \times N_A = \dfrac{1000}{17.13} \times 6.0 \times 10^{23} = \dfrac{6}{17.13} \times 10^{26}$
Total energy released ($E$) is:
$E = N \times Q = \left(\dfrac{6}{17.13} \times 10^{26}\right) \times (17.0373 \times 10^6 \text{ eV})$
$E = \dfrac{102.2238}{17.13} \times 10^{32} \text{ eV} \approx 5.967 \times 10^{32} \text{ eV}$
Rounding to the nearest integer, we get $\alpha = 6$.
Answer: $6$
Hence, the correct answer is 6.