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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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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