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In the investigation of the alpha-decay of polonium, alpha-particles with energies of 5.30 and 4.50 MeV were detected. Find the energy of the gamma-rays emitted in the decay, taking account of the recoil of the nucleus. |
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Answer» `epsi_(gamma)=5.30-4.50=0.80MeV`. HOWEVER, this estimate is too rough, since it takes no account of the recoil energy of the nucleus. Certain problems in nuclear physics demand a much higher accuracy. Consider the energy LEVEL DIAGRAM for the case of polonium decay. The letter `Pb^(**)` with an asterisk denotes the excited lead nucleus which emits gamma-photons. The total energy of the transition is `epsi_(1)=K_(1alpha)+R_(1),"where "K_(1alpha)=5.30MeVandR_(1)` is the recoil energy of the nucleus. Since `R_(1)-4K_(1alpha)//206`, it follows that `epsi_(1)=5.30+(5.30xx4//206)=5.40MeV` Similarly, the energy liberated in the course of the second transition is `epsi_(2)=4.50+(4.50xx4//206)=4.58MeV` The energy of the gamma-photon is `epsi_(gamma)=epsi_(1)-epsi_(2)=5.40-4.58=0.82MeV` Note that because the mass of the lead nucleus is large, we can neglect the recoil energy resulting from the emission of a gamma-photon. 
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