The only source of energy in a particular star is the fusion reaction given by -
`3._(2)He^(4)to._(6)C^(12)`+energy
Masses of `._(2)He^(4)` and `._(6)C^(12)` are given
`(m._(2)He^(4))=4.0025 u, m(._(6)C^(12))=12.0000u`
Speed of light in vaccume is `3xx10^(8) m//s`. power output of star is `4.5xx10^(27)` watt. The rate at which the star burns 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`