The masses of ¹¹C and ¹¹B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the β*-decay of ¹¹C to ¹¹B.

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Solution

Given:
Mass of ¹¹C, m(¹¹C) = 11.0114 u
Mass of ¹¹B, m(¹¹B) = 11.0093 u
Energy liberated in the β⁺ decay (Q) is given by
$Q = [m ({}^{11}C) - m ({}^{11}B) - 2 m_{e}] c^{2}$
= (11.0114 u − 11.0093 u $-$ 2 $\times$ 0.0005486 u)c²
= 0.0010028 $\times$ 931 MeV
= 0.9336 MeV = 933.6 keV
For maximum KE of the positron, energy of neutrino can be taken as zero.
∴ Maximum KE of the positron = 933.6 keV

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Similar questions

The masses of $^{11} C$ and $^{11} B$ are respectively 11.0114 u and 11.0093 u. Find the maximum energy a position can have in the $β^{+} -$ decay of $^{11} C$ to $^{11} B$

^{11} C

undergoes

β^{+}

decay to

^{11} B .

What is the maximum energy a positron emitted in the process can possess?
Atomic masses:

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

^{11} C u n d e r g o e s β^{+}

decay to

^{11} B .

What is the maximum energy a positron emmited in the process can possess?
Atomic masses:

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

^{11} C

undergoes

β^{+}

decay to

^{11} B .

What is the maximum energy a positron emitted in the process can possess?
Atomic masses:

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

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

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The masses of 11C and 11B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the β*-decay of 11C to 11B.

The masses of ¹¹C and ¹¹B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the β*-decay of ¹¹C to ¹¹B.