The radionuclide $^{56} M n$ has a half-life of 2.58 h and is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope $^{55} M n$ , and the manganesedeuteron reaction that produces $^{56} M n$ is
$^{55} M n + d \to^{56} M n + p$
If the bombardment lasts much longer than the half-life of $^{56} M n$ , the activity of the $^{56} M n$ produced in the target reaches a final value of $8.88 \times 1 0^{1} 0 B q$ . (a) At what rate is $^{56} M n$ being produced? (b) How many $^{56}$ Mn nuclei are then in the target? (c) What is their total mass?

Question

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Answer 1

If $M_{H} e$ is the mass of an atom of helium and $M_{C}$ is the mass of an atom of carbon, then
the energy released in a single fusion event is
$Q = (3 M_{H} e - M c) c^{2} = 3 [3 (4.0026 u) (12.0000 u)] (931.5 M e V / u) = 7.27 M e V$ .
Note that $3 M_{H} e$ contains the mass of six electrons and so does $M_{C}$ . The electron masses
cancel and the mass difference calculated is the same as the mass difference of the nuclei.

Answer 2

If

M_{H} e

is the mass of an atom of helium and

M_{C}

is the mass of an atom of carbon, then
the energy released in a single fusion event is

Q = (3 M_{H} e - M c) c^{2} = 3 [3 (4.0026 u) (12.0000 u)] (931.5 M e V / u) = 7.27 M e V

.
Note that

3 M_{H} e

contains the mass of six electrons and so does

M_{C}

. The electron masses
cancel and the mass difference calculated is the same as the mass difference of the nuclei.