Question

In nuclear fusion, $1gm$ hydrogen is converted into $0.993gm$ Helium in one hour. If the efficiency of the generator is $5$%, the energy obtained in $KWh$ is:

• A. $8.75\times {10}^3$
• B. $4.75\times {10}^3$
• C. $5.75\times {10}^3$
• D. $3.73\times {10}^3$

Answer 1

Answer: (A)

$8.75\times {10}^3$

Mass defect $=(1-0.993)gm=7\times {10}^{-3}gm$
From energy mass equivalence,
$E=m{c}^2$
$=7\times {10}^{-6}\times (3\times {10}^8{)}^{2}joules$
$=6.3\times {10}^{11}J$

Efficiency $=5$%
Energy obtained $=3.15\times {10}^{10}joules$
and $1KWh=1000\times 3600joules=3.6\times {10}^{6}joules$
Amount of energy tapped $=\frac{3.15\times {10}^{10}}{3.6\times {10}^6}KWh$

$=8.75\times {10}^{3}KWh$