Question

If the atomic masses of lithium, helium and proton are 7.01823 amu, 4.00387 amu and 1.00815 amu respectively, calculate the energy that will be evolved in the reaction. $\mathbf{L}\mathbf{i}^{\mathbf{7}}\mathbf{+}\mathbf{H}^{\mathbf{1}}\mathbf{\rightarrow}\mathbf{2H}\mathbf{e}^{\mathbf{4}}\mathbf{+}\text{energy}$.

(Given that 1 amu = 931 MeV)

• A. 17.3 MeV
• B. 17.8 MeV
• C. 17.2 MeV
• D. 17.0 MeV

Answer 1

Answer: (A)

17.3 MeV

Answer 2

Total mass of the reacting species

$(Li^{7}$and$H^{1}) = 7.01823 + 1.00815 = 8.02638amu$

The mass of the resulting species

$(2He^{4}) = 2 \times 4.00387 = 8.00774$ amu

Mass of reacting species converted into energy, i.e., $\Delta m = 8.02638 - 8.00774 = 0.01864$amu

∴ Energy evolved in the reaction

$= 0.01864 \times 931.5 = 17.363$MeV.