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Alpha Decay

Write symboli...

Question

Write symbolically the nuclear $β^{+}$ decay process of $_{6}^{11} C$ . Is the decayed product X an isotope or isobar of ( $_{6}^{11} C$ )? Given the mass values m ( $_{6}^{11} C$ ) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process.

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Solution

$β^{+}$ decay:

$_{116} C \to_{115} X + e^{+} + ν$

$e^{+} =$ Positron

$ν =$ Neutrino

As mass number of $C$ and $X$ are the same, $C$ and $X$ are isobars.

Energy released $Q = (m_{c} - m_{x} - m_{e}) c^{2}$

$= [(m_{c} + 116 m_{e}) - (m_{x} + 115 m_{e})] c^{2}$

$= [m_{c} - m_{x}] c^{2}$

$= [11.011434 - 11.009305] c^{2}$

We have $c^{2} = 931.5 M e V / a m u$

$= 0.002129 \times 931.5 M e V$

$= 1.983$ $M e V$

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_{6}^{11} C

_{6}^{11} C

N = N 0 e^{- λ} t

β +

_{11}^{22} N a

_{11}^{22} N a

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

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m_{C} = 11.011433 u, m_{e} = 0.0005486 u, m_{B} = 11.009305 u, 1 a m u = 931 M e V

\frac{100 Q}{19}

^{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

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