Q. In nuclear fusion, One gram hydrogen is converted into 0.993gm. If the efficiency of the generator be 5%, then the energy obtained in KWH is

(a) 8.75 ×10³

(b) 4.75 × 10³

(c) 5.75 × 10³

(d) 3.73 × 10³

Ans: (a)

Sol: Δm = 1−0.993 = 0.007g =7×10^{-6} kg

E = Δmc^{2}

= (7×10^{-6})(3×10^{8})^{2}

= 63 × 10^{-10} J

Energy required = (5/100)×63 × 10^{-10}

= 63 × 5× 10^{-8}

Since , 1KWH = 3.6× 10^{6} J

Hence, Energy obtained = (63 × 5× 10^{-8})/(3.6× 10^{6} )

= 8.75 ×10³ KWH