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

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Modern Physics

The isotope

_{5}^{12} B

having a mass 12.014 u undergoes

β

-decay to

_{6}^{12} C

_{6}^{12} C

has an excited state of the nucleus

(_{6}^{12} C^{*})

at 4.041 MeV above its ground state. If

_{5}^{12} B

decays to

_{6}^{12} 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).

Answer: 9

β^{-}

decay the energy released, expressed with atomic masses, is

Q = [M (^{12} B) - M (^{12} C^{*})] c^{2}

, where the excited carbon lies

4.041

MeV above the ground state whose atomic mass is exactly

12.000

u. Thus

Q = (12.014 - 12.000) (931.5) - 4.041 = 13.041 - 4.041 = 9.000

MeV. The

β

particle carries its maximum kinetic energy when the antineutrino takes essentially none of the energy, so the maximum kinetic energy is

9

MeV.

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Integer answer