The loss in mass when 1 gm of ${ }_{3} \mathrm{Li}^{7}$ is converted into two ${ }_{2} \mathrm{He}^{

Question

The loss in mass when 1 gm of ${ }_{3} \mathrm{Li}^{7}$ is converted into two ${ }_{2} \mathrm{He}^{4}$ by proton bombardment is\ (mass of ${ }_{3} \mathrm{Li}^{7}=7.0188 \mathrm{amu}$ )\ (mass of ${ }_{2} \mathrm{He}^{4}=4.0040 \mathrm{amu}$ )\ (mass of proton $=1.0081 \mathrm{amu}$ )\

Professor PC Thomas Classes · Answer

${ }_{3} \mathrm{Li}^{7}+{ }_{1} \mathrm{H}^{1} ightarrow 2{ }_{2} \mathrm{He}^{4}$ \ Loss of mass $=(7.0188+1.0081)-$\ $2 	imes 4.0040 \mathrm{amu}=0.0189 \mathrm{amu}$\ This loss is for 7 amu\ $	herefore$ Loss for 7 gram $=0.0189 \mathrm{~g}$\ Loss for 1 gram $=\frac{0.0189}{7}=0.0027$ gram\