The isotope 52 B having a mass 12-014 u undergoes -decay to 612 C -62 C has an excited state of the nucleus 52 C - at 4-041 MeV above its ground state- If 12-5 B decays to 612 C - the maximum kinetic energy of the -particle in units of MeV is 1 u -931-5 MeV - c 2- where c is the speed of light in vacuum-

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Standard XII

Physics

Conservation of Momentum

The isotope 5...

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 ( $_{5}^{12} C^{}$ ) at $4.041 M e V$ above its ground state. If $_{5}^{12} B$ decays to $_{6}^{12} C^{}$ , the maximum kinetic energy of the $β$ -particle in units of $M e V$ is
( $1 u = 931.5 M e V / c^{2}$ , where c is the speed of light in vacuum).

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Solution

$_{5}^{12} B \to_{6}^{12} C + β^{-} + ¯ ν$
For any radioactive particle decay, the energy released in the reaction will be given by $Q = (m_{B} - m_{C}) c^{2}$
Now $m_{C} = 12 u$ and $m_{B} = 12.041 u$
Out of $Q$ , 4.041 eV energy is used to excite $_{6}^{12} C$ to its excited state $_{6}^{12} C^{*}$ , then K.E. available to $β$ particle and antineutrino is $= (m_{B} - m_{C}) c^{2} - Δ m c^{2} = 0.014 \times 931.5 - 4.041 = 9 M e V$
The $β$ particle will have maximum K.E when antineutrino will have minimum K.E i.e 0.
Therefore maximum K.E of the $β$ particle is $9 M e V$

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Q. 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 (

_{5}^{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).

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 $_{5}^{12} B$ decay to $_{6}^{12} C^{*},$ the maximum kinetic energy of the $β -$ particle in units of $MeV$ nearly is (integer only).

Take : $m (_{6}^{12} C) = 12.0 u (1 u = 931.5 MeV / c^{2}, Where c is the speed of light in vacuum)$

Q. 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 (

_{5}^{12} C^{*}

) at

4.041 M e V

above its ground state. If

_{5}^{12} B

decays to

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

, the maximum kinetic energy of the

β

-particle in units of

M e V

is
(

1 u = 931.5 M e V / c^{2}

, where c is the speed of light in vacuum).

Suppose a $_{88}^{226} R a$ nucleus at rest and in ground state undergoes $α$ -decay to a $_{86}^{222} R n$ nucleus in its excited state. The kinetic energy of the emitted $α$ particle is found to be $4.44 M e V$ . $_{86}^{222} R n$ nucleus then goes to its ground state by $γ$ -decay. The energy of the emitted $γ$ photon is $k e V$ .

[Given : atomic mass of $_{88}^{226} R a = 226.005 u$ , atomic mass of $_{86}^{222} R n = 222.000 u$ , atomic mass of $α$ particle = $4.000 u, 1 u = 931 M e V / c^{2}, c$ is speed of light]

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