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3. Ra-226 (226.025403 u) decays spontaneously by alpha' emission to Rn-222 (222.017571 u): 220Ra → 222Rn+a+Q. Calculate: Z 88 a) the total energy released by the process; b) the kinetic energy of alpha particle. a: 4.002603 u; u 931.5MeV 2 zh

3. Ra-226 (226.025403 u) decays spontaneously by alpha' emission to Rn-222 (222.017571 u): 220Ra → 222Rn+a+Q. Calculate: Z 88 a) the total energy released by the process; b) the kinetic energy of alpha particle. a: 4.002603 u; u 931.5MeV 2 zh

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3. Ra-226 (226.025403 u) decays spontaneously by alpha emission to Rn-222 (222.017571 u): 222 Ra → 2² Rn +α + Q. Calculate: Z 226 88 a) the total energy released by the process; b) the kinetic energy of alpha particle. a) Q = (m 1) According Ассалі воспи 2.Q - m Ra M.. -- 4.roy) to the law of momentum conservations: Rn 226,0253. Accordil to the bur m MRIA a: 4.002603 u; u 931.5MeV m₂) = Urim + m; Ma I the law of KE coverch 2 MRA VRn + m₂ "x тка и 34 zh 국 4 -25 -th
Transcribed Image Text:3. Ra-226 (226.025403 u) decays spontaneously by alpha emission to Rn-222 (222.017571 u): 222 Ra → 2² Rn +α + Q. Calculate: Z 226 88 a) the total energy released by the process; b) the kinetic energy of alpha particle. a) Q = (m 1) According Ассалі воспи 2.Q - m Ra M.. -- 4.roy) to the law of momentum conservations: Rn 226,0253. Accordil to the bur m MRIA a: 4.002603 u; u 931.5MeV m₂) = Urim + m; Ma I the law of KE coverch 2 MRA VRn + m₂ "x тка и 34 zh 국 4 -25 -th
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