Search any question & find its solution
Question:
Answered & Verified by Expert
In the following equation representing $\beta^{-}$ decay, the number of neutrons in the nucleus $X$ is
${ }_{83}^{210} \mathrm{Bi} \longrightarrow X+e^{-1}+\bar{v}$
Options:
${ }_{83}^{210} \mathrm{Bi} \longrightarrow X+e^{-1}+\bar{v}$
Solution:
2118 Upvotes
Verified Answer
The correct answer is:
$126$
In $\beta^{-}$decay, reaction is given as,
${ }_{83}^{210} \mathrm{Bi} \longrightarrow{ }_{84}^{210} X+e^{-1}+\overline{\mathrm{v}}$
$\therefore$ Number of neutrons $=210-84=126$
${ }_{83}^{210} \mathrm{Bi} \longrightarrow{ }_{84}^{210} X+e^{-1}+\overline{\mathrm{v}}$
$\therefore$ Number of neutrons $=210-84=126$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.