an injection containing 20.0 pg 10 5B was injected into a patient suffering from

Question

an injection containing 20.0 pg 10 5B was injected into a patient suffering from a brain tumor. The patient was irradiated by 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 release, consider the isotopic masses as follows 10 5B 10.0129 amu Neutron 1.0086 amu 7 3Li 7.0160 amu Alpha particles 4.0026 amu Express answer in megalectronvolts to three significant figures Energy released=?? MeV

Explanation / Answer

5B10 + 0n1 ----------------> 3Li7+ 2He4

For 5B10

no of protons in 5B10 = 5

no. of neutrons = 10- 5 = 5

mass of proton = 1.00728 amu

mass of 6 protons = 5 x 1.00728 = 5.0364 amu

mass of neutron = 1.00867 amu

mass of 10 neutrons = 5 x 1.00867 = 5.04335 amu

total mass of 5B10    = 5 protons mass + 5 neutrons mass

                                = 5.0364 amu + 5.04335 amu

                                 = 10.07975 amu

Actual mass of 5B10 = 10.0129 amu.

mass defect = calculated mass - actual mass

                    = 10.07975 amu - 10.0129 amu.

                    = 0.06685 amu

1 amu = 931.5 MeV

0.06685 amu = 0.06685 x 931.5 MeV = 62.27 MeV

Mass defect = 62.27 MeV

Binding Energy = mass defect in amu x 931.5 MeV

                         = 0.06685 x 931.5 MeV

                         = 62.3 MeV

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