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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Answer 1

$β^{+}$ 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 $c^{2} = 931.5 M e V / a m u$

$= 0.002129 \times 931.5 M e V$

$= 1.983$ $M e V$

Answer 2

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