The only source of energy in a particular star is the fusion reaction as given below :
`3 " " 2^(He^(2))rarr6^(C^12)+"energy"`
The masses of `2^(He^(4))` and `6^(C^(12))` are given as
`m(2^(He^(4)))=4.0025mu`
`m(6^(C^(12)))=12.000mu`
The speed of light in vacuum `=3xx10^(8)` m/s if power output of the star is `4.5xx10^(27)` W. The rate (in kg /s) at which the star "bunrs" helium is

Options

A. `8xx10^(12) kg//s`
B. `4xx10^(13) kg//s`
C. `8xx10^(13) kg//s`
D. `6xx10^(13) kg//s`

Explanation

Fraction of mass converted in energy
`(3xx4.0025-12.0000)/(3xx4.0025)=0.0075/12-("Rate of loss of mass")/("Rate of burning")`
Rate of burnig `=12/(75xx10^(-4))xx("Power output")/(C^(2))`
`=12/(775xx9xx10^(12))=54/(9xx75)xx10^(15)=2/25xx10^(15)=8xx10^(13) kg//s`