Question

An injection containing 13.0 pg 10 5B was injected into a patient suffering from a brain tumor. The patient was irradiated by a neutron gun at the site of the tumor. Calculate the amount of energy released in megaelectronvolts given that 931 MeV is released for each 1 amu mass defect. To calculate the energy released, consider the isotopic masses as follows: Element or particle Isotopic mass (amu) 10 5 B 10.0129 neutron 1.0086 7 3Li 7.0160 alpha particle 4.0026

Answer 1

The reaction that occurs is
¹⁰₅B + ¹₀n →⁷₃Li + ⁴₂α + energy

Step 1. Find the energy released from the reaction for a single ¹⁰₅B atom.

The energy (E=mc²) is due the mass loss (mass defect) m.

m = (mass of ¹⁰₅B + ¹₀n) - (mass of ⁷₃Li + ⁴₂α)
= (10.0129 + 1.0086) - (7.0160 + 4.0026)
= 0.0029 amu

Since 1amu is equivalent to an energy of 931 MeV
E = 0.0029 x 931 = 2.70 MeV
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Step 2 Find how many ¹⁰₅B atoms there are

13.0pg = 13.0x10⁻¹² g = 13.0x10⁻¹⁵ kg

1amu = 1.66x10⁻²⁷ kg
Mass of 1 atom of ¹⁰₅B = 10.0129 amu = 10.0129 x 1.66x10⁻²⁷ = 1.662x10⁻²⁶ kg

No of atoms of ¹⁰₅B = 13.0x10⁻¹⁵ / (1.662x10⁻²⁶) = 0.782x10¹²
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Step 3 Find the total energy
Total energy = 0.782x10¹² x 2.70MeV = 2.111x10¹² MeV