Atoms & Nuclei

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180 of 239

Question 180 of 239

The energy released when 717.13\dfrac{7}{17.13} kg of 37Li^{7}_{3}\text{Li} is converted into 24He^{4}_{2}\text{He} by proton bombardment is α×1032\alpha \times 10^{32} eV. The value of α\alpha is _______. (Nearest integer) (Mass of 37Li=7.0183^{7}_{3}\text{Li} = 7.0183 u, mass of 24He=4.004^{4}_{2}\text{He} = 4.004 u, mass of proton =1.008= 1.008 u and 11 u =931= 931 MeV/c2^2 and Avogadro number =6.0×1023= 6.0 \times 10^{23})

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Answer: 6

The nuclear reaction for the proton bombardment of Lithium is:

37Li+11H→224He^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow 2 ^{4}_{2}\text{He}

The mass defect (Δm\Delta m) for the reaction is:

Δm=m(37Li)+m(11H)−2m(24He)\Delta m = m(^{7}_{3}\text{Li}) + m(^{1}_{1}\text{H}) - 2m(^{4}_{2}\text{He})

Δm=7.0183+1.008−2(4.004)\Delta m = 7.0183 + 1.008 - 2(4.004)

Δm=8.0263−8.008=0.0183 u\Delta m = 8.0263 - 8.008 = 0.0183 \text{ u}

Energy released per reaction (QQ) is:

Q=Δm×931 MeV=0.0183×931 MeV=17.0373 MeVQ = \Delta m \times 931 \text{ MeV} = 0.0183 \times 931 \text{ MeV} = 17.0373 \text{ MeV}

Number of moles of 37Li^{7}_{3}\text{Li} in 717.13 kg\dfrac{7}{17.13} \text{ kg}:

n=717.13×103 g7 g/mol=100017.13 moln = \dfrac{\dfrac{7}{17.13} \times 10^3 \text{ g}}{7 \text{ g/mol}} = \dfrac{1000}{17.13} \text{ mol}

Number of atoms of 37Li^{7}_{3}\text{Li}:

N=n×NA=100017.13×6.0×1023=617.13×1026N = n \times N_A = \dfrac{1000}{17.13} \times 6.0 \times 10^{23} = \dfrac{6}{17.13} \times 10^{26}

Total energy released (EE) is:

E=N×Q=(617.13×1026)×(17.0373×106 eV)E = N \times Q = \left(\dfrac{6}{17.13} \times 10^{26}\right) \times (17.0373 \times 10^6 \text{ eV})

E=102.223817.13×1032 eV≈5.967×1032 eVE = \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 α=6\alpha = 6.

Answer: 66

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