The sun has a power of P=3.85`xx10^(26)` W and the only source of energy in it is the following reaction
`4^(1)Hrarr.^(4)He+2e^(+)+2v_(0)`
The 'electron neutrinos `(V_(e))'` are nearly mass less and carry negligible energy. However, they are able to escape from the sun and have been detected on the earth. The Earth-sun distance is `r=1.5xx10^(11)m` and masses of a hydrogen ato, helium atom and a positron are `1.6740xx10^(-27)kg,6.6450xx10^(-27)kg` and `0.0009xx10^(-27)kg` respectively.
(a) Calculate the flux density (i.e. number of neutrinos arriving at the Earth) in units of `m^(-2)s^(-1)`.
(b) While travelling from the Sun to the Earth, some of the electron neutrinos `(V_(e))` are converted into other types of neutrinos `-v_(0)`. the dectector on the Earth has `(1)/(5)` th efficiency for detecting `v_(0)` as compared to its efficiency to detect `v_(0)` Had there been no conversions of `v_(e)`. we expect to detect `N_(1)` neutrinos in a year. However, due to conversion, we detect only `N_(2)` neutrinos (`v_(0)` and `v_(e)` combined) per year. what fraction (f) of `v_(e)` gets convertedinto `v_(0)`. express your answer in terms of `N_(1)` and `N_(2)`.

Answer

(a) `5.9xx10^(14)m^(-2)s^(-1)` (b) `(5)/(4)(1-(N_(2))/(N_(1)))`