Before the discovery of the neutron, it was proposed that the penetrating radiation produced when beryllium was bombarded with alpha particles consisted of high energy gamma rays (up to 50 MeV) produced in reactions such as alpha+9 Be--->13 C+ gamma.a). Calculate the Q-value for this reaction. b). If 5 MeV alpha particles are incident on 9Be, calculate the energy f the 13C nucleus and hence, determine the energy of gamma radiation assuming it is emitted as a single photon. Hint: you may neglect the momentum of the gamma ray relative to the 13C nucleus. Masses: m (4He)= 4.0026u, m(9Be)= 9.0122u, m (13C)= 13.0034 u.

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Answer:

Answer part a:

Q= 10.62 MeV

Answer part b:

Energy of gamma particle is given by

E γ = 10.62 MeV

Explanation:

See attached file for explanation.

Answer:

a. 0.00194 u b. 16.25 eV; 2.899 X 10^-13 J of energy released

Explanation:

4He + 9 Be -> 13 C + n

13.034-(4.0024+9.0122)=0.0194 u, 1 u = 1.66054 x10^-27

E=Δmc^2

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