The kinetic energy of x-particle emitted...

The kinetic energy of x-particle emitted in the `alpha`-decay of `"_88Ra^226` is [Given, mass number Ra = 222u]

5.201 MeV

3,301 MeV

6.023 MeV

4.871 MeV

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Find the Q-value and the kinetic energy of the emitted alpha -particle in the alpha -decya of (a) ""_(88)^(226)Ra and (b) ""_(86)^(220)Rn . Given m (""_(88)^(226)Ra) = 226.02540 u, " " m(""_(86)^(222)Rn) = 222.01750 u . m(""_(86)^(220)Rn) = 220.01137 u, " " m (""_(84)^(216)Po) = 216.00189u .

Th^228 emits an alpha particle to reduce to Ra^224 .Calculate the kinetic energy of the alpha particle emitted in the following decay: Th^228 rarr Ra^**+ alpha (Ra^224)^** rarr Ra^224+ gamma(217 keV) Atomic mass of Th^228 is 228.028726 u ,that of Ra^224 is 224.020196 u and that of He_2^4 is 4.00260 u.

Find the kinetic energy of the alpha - particle emitted in the decay ^238 Pu rarr ^234 U + alpha . The atomic masses needed are as following: ^238 Pu 238.04955 u ^234 U 234.04095 u ^4 He 4.002603 u . Neglect any recoil of the residual nucleus.

Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme: N^12 rarr C^12 + e^(+)+v C^12 rarr C^12 + gamma(4.43 MeV). The atomic mass of N^12 is 12.018612 u.

Consider the beta decay ^198 Au rarr ^198 Hg ** + Beta^(-1) + vec v . where ^198 Hg^** represents a mercury nucleus in an excited state at energy 1.088 MeV above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic mass of ^198 Au is 197.968233 u and that of ^198 Hg is 197.966760 u .

The nucleus ""_(10)^(23)Ne decays by beta^(-) emission. Write down the beta -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that : m (""_(10)^(23) Ne) = 22.994466 u m (""_(11)^(23)Na) = 22.989770 u .

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