The isotope having a mass undergoes -decay to . has an excited state of the nucleus at above its ground state. If decays to , the maximum kinetic energy of the -particle in units of MeV is (, where is the speed of light in vacuum). [2016]

Question

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). [2016]

Smart Achievers · Accepted Answer

(9)

Here, ${}_{5}^{12}β ⟶ {}_{6}^{12}C + {}_{- 1}^{0}e + \bar{ν}$

Maximum kinetic energy of $({{}_{6}^{12}C}^{*})$ -particle

$= [mass of {}_{5}^{12}β - mass of {}_{6}^{12}C] \times 931.5 - 4.041$

$= [12.014 - 12] \times 931.5 - 4.041 = 9 MeV$

Solution

Ans.
(9)

Here, ${}_{5}^{12}β ⟶ {}_{6}^{12}C + {}_{- 1}^{0}e + \bar{ν}$

Maximum kinetic energy of $β$ -particle

$= [mass of {}_{5}^{12}β - mass of {}_{6}^{12}C] \times 931.5 - 4.041$

$= [12.014 - 12] \times 931.5 - 4.041 = 9 MeV$

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (9) Here, β512⟶C612+e-10+ν¯ Maximum kinetic energy of β-particle =[mass of β512-mass of C612]×931.5-4.041 =[12.014-12]×931.5-4.041=9 MeV

Want Us to Email you About Special Offers & Updates?

Ans.
(9)

Here, ${}_{5}^{12}β ⟶ {}_{6}^{12}C + {}_{- 1}^{0}e + \bar{ν}$

Maximum kinetic energy of $β$ -particle

$= [mass of {}_{5}^{12}β - mass of {}_{6}^{12}C] \times 931.5 - 4.041$

$= [12.014 - 12] \times 931.5 - 4.041 = 9 MeV$