Question

${}_{3}L{i}^7{+}_1{H}^2{\to }_{4}B{e}^8{+}_o{n}^1+Q$
Mass of ${}_{3}L{i}^7=$ $7.01823amu$
Mass of ${}_1{H}^2=$ $2.01474amu$
Mass of ${}_{4}B{e}^8=$$8.00785amu$
Mass of ${}_o{n}^1=$ $1.00899amu$
Then, the value of Q is

• A. $5MeV$
• B. $10MeV$
• C. $15MeV$
• D. $0$

Answer 1

Answer: (C)

$15MeV$

Mass of reactants is: $7.01824+2.01474=9.03297amu$
Mass of products is: $8.00785+1.00899=9.01684amu$
The difference in mass $=0.01613amu$. (This mass has been converted to energy).
$1amu=1.67\times {10}^{-27}kg$
Hence, mass difference is $=0.01613\times 1.67\times {10}^{-27}kg=2.69371\times {10}^{-29}kg$
Using Einstein's relation:
$E=m{c}^2$
$E=2.69371\times {10}^{-21}\times (3\times {10}^8{)}^2=2.42\times {10}^{-12}J$
$E=15MeV$