The mass defect in a particular nuclear reaction is `0.3` grams. The amont of energy liberated in kilowatt hours is.
(Velocity of light `= 3 xx 10^8 m//s`).

Options

A. `7.5xx10^6`
B. `3xx10^6`
C. `1.5xx10^6`
D. `2.5xx10^6`

Answer

A

Explanation

`E= (Deltam)c^2 = (0.3)/(1000) xx 3xx10^8 xx 3 xx 10^8`
`=2.7xx10^13 J`
1KW hour `=10^3 J//s xx 60 xx 60s = 3.6xx 10^6 J`
`therefore E=(2.7xx10^13)/(3.6xx10^6) = 3/4 xx10^7 = 7.5xx10^6 KWh`