The sun radiates energy at the rate of `4xx10^(26)` joule `sec^(-1)`. If the energy of fusion process
`4 ._(1)^(1)H rarr ._(2)^(4)He + 2 ._(1)^(0)e`
is `27MeV`, calculated amount of hydrogen atoms that would be consumed per day for the given process.

Explanation

27 MeV `= 27 xx 10^(6) xx 1.6 xx 10^(-19)`
`= 43.2 xx 10^(-13) J`
Energy radiated by the sum per day.
`= 4 xx 10^(26) xx 3600 xx 24 J "day"^(-1)`
`= 34.56 xx 10^(30) J "day"^(-1)`
`43.2 xx 10^(-13)` J of energy is obtained from
= 4 amu of H
`= 4 xx 1.66 xx 10^(-24) g` of H
`34.56 xx 10^(30) J` of energy is obtained from
`= (4 xx 1.66 xx 10^(24))/(43.2 xx 10^(-13)) xx 34.56 xx 10^(30)`
`= 5.31 xx 10^(19) g`