In a slow reaction, heat is being evolved at a rate about 10 m W in a liquid. If the heat were being generated by the decay `of ^(32)` P, a radioactive isotope of phosphorus that has half-life of 14 days and emits only beta-particles with a mean energy of 700KeV, estimate the number `of ^(32)` P atoms in the liquid. Express your answer in form of `Axx10^(15)` and fill A in OMR sheet. Round off A to nearest integer [Take:l n`2=0.7`]

Answer

`0154`

Explanation

`P=700xx10^(3)xx1.6xx10^(-19)xx(dN)/(dt)=10xx10^(-3)`
`(dN)/(dt)=10^(-2)/10^(-14)xx1/(7xx16)=10^(12)/11.2=lambda N_(0)`
`lambda=(l n2)/(14xx86400) Rightarrow N_(0)=(14xx86400xx10^(12))/(11.2l n2)=154xx10^(15)`