The radionuclide 11C decays according to11 11 +6 5 1/2 C B+ + : =20.3 min → e T νThe maximum energy of the emitted positron is 0.960 MeV.Given the mass values:m (116C) = 11.011434 u and m (116B ) = 11.009305 u,calculate Q and compare it with the maximum energy of the positron emitted.

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Solution

Given: The mass of C 6 11 is 11.011434 u, the mass of B 6 11 is 11.009305 u, the half life of the U 92 235 is 20.3 minand the maximum energy of emitted positron is 0.960 MeV.

The given nuclear reaction is,

C 6 11 → B 6 11 + e + +v

The Q-value of the reaction is given by,

Q-value=( m i − m f ) c 2 =[ m( C 6 11 )−{ m( B 6 11 )+2 m e } ] c 2

Where, the sum of initial mass is m i and the sum of final mass is m f .

By substituting the given values in above equation, we get

Q-value=[ 11.011434−11.009305−2×0.000548 ] c 2 =( 0.001033 c 2 ) u

We know that,

1 u=931.5 MeV/ c 2

Thus, the Q value is,

Q-value=0.001033×931.5 ≈0.962 MeV

The Q-valueis almost equal to the maximum energy of the emitted positron.

Thus, the Q-value for positron emitted is 0.962 MeV.

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Similar questions

^{11} C

_{6}^{11} C \to_{5}^{11} B + e^{+} + ν

T_{1 / 2}

= 20.3 m i n

0.960 M e V

m (_{6}^{11} C) = 11.011434 u

m (_{6}^{11} B) = 11.009305 u

Q

The radionuclide ¹¹C decays according to

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

calculate Q and compare it with the maximum energy of the positron emitted

^{11} C

β^{+}

^{11} B .

^{11} C : 11.0114 u;^{11} B : 11.0093 u; e^{-} : 0.005486. c^{2} = 931 \frac{M e V}{u}

^{11} C

β^{+}

^{11} B .

^{11} C : 11.0114 u;^{11} B : 11.0093 u; e^{-} : 0.005486. c^{2} = 931 \frac{M e V}{u}

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