1. Tardigrade
  2. Question
  3. Physics
  4. The isotope 512 B having a mass 12.014 u undergoes β-decay to 612 C . 612 C has an excited state of the nucleus ( 612 C *. ) at 4.041 MeV above its ground state. If 512 B decays to 612 C *, the maximum kinetic energy of the β-particle in units of MeV is. (1 u =931.5 MeV / c 2., where c is the speed of light in vacuum).

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The isotope ${ }_{5}^{12} B$ having a mass $12.014 u$ undergoes $\beta$-decay to ${ }_{6}^{12} C .{ }_{6}^{12} C$ has an excited state of the nucleus $\left({ }_{6}^{12} C ^{*}\right.$ ) at $4.041 \,MeV$ above its ground state. If ${ }_{5}^{12} B$ decays to ${ }_{6}^{12} C ^{*}$, the maximum kinetic energy of the $\beta$-particle in units of $MeV$ is. $\left(1 u =931.5 \,MeV / c ^{2}\right.$, where $c$ is the speed of light in vacuum).

JEE AdvancedJEE Advanced 2016

Solution:

${ }_{5}^{12} B \longrightarrow { }_{6}^{12} C +{ }_{-1}^{0} \beta$
Mass loss $=0.014\, u$
Energy produced $=13.041\, MeV \left( E = mc ^{2}\right)$
$\therefore $ Maximum Kinetic energy of $\beta$-particle
$=13.041-4.041=9 \,MeV$