The radionuclide 11 C decays according to 11 C 11 B e v, T 1 2 20 3 The maximum energy of the positions is 0 960MeV (a) Show that the designation energy Q for this process is given by Q (m C m B 2me) c 2 where m C and m B are the atomic masses of 11 C and 11 B, respectively, and me is the mass of a positron (b) Given the mass values m C 11 011 424 u, m B 11 009 305 u, and m e 0 000 548 6 u, calculate Q and compare it with the maximum energy of the emitted positron given above

Question: The radionuclide 11 C decays according to 11 C 11 B + e + + v, T 1/2 = 20.3. The maximum energy of

The radionuclide ¹¹C decays according to ¹¹C → ¹¹B + e⁺ + v, T_1/2 = 20.3. The maximum energy of the positions is 0.960MeV.

(a) Show that the designation energy Q for this process is given by Q = (m_C - m_B – 2me) c² where m_C and m_B are the atomic masses of ¹¹C and ¹¹B, respectively, and me is the mass of a positron.

(b) Given the mass values m_C = 11.011 424 u, m_B = 11.009 305 u, and m_e= 0.000 548 6 u, calculate Q and compare it with the maximum energy of the emitted positron given above.

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