Complete the following radioactive decay problem. 252/98 cf -> ? + 4/2 he
14N7 + 1n0 = 14C6 + 1H1 Explanation: The question shows how atoms are being represented by the number of proton and the number of electron. By simply taking mass balance for the left and right hand sides, that is: 14N+1n =?C+ 1H ?C = 15-1 =14 Also, N7+n0 = C? + H1 C? = 7+0-1 C? =6 Hence, ?C? = 14C6
238/92 U → 4/2 He + 234/90 Th → 234/91 Pa + 0/-1e Explanation: The first decay is considered as alpha decay 238/92 U → 4/2 He + 234/90 Th.The second decay of 234/90 Th is considered to be a beta decay that the change is in the atomic number from 90 to 91 so it losses 1 electron.So, the second decay will be: 234/90 Th → 234/91 Pa + 0/-1 e
230 90Th Explanation: A careful observation of the equation given in question shows that 234 92U is undergoing alpha decay. This means that the resulting daughter nuclei will have a decrease of 4 in the mass number and a decrease of 2 in the atomic number. Please see attached photo for further details.
238 94Pu Explanation: From the question given, we noticed that the element is undergoing beta minus decay. This means that the daughter nuclei obtained will have the same mass number as the parent element and the the atomic number of the daughter nuclei will increase by 1. Please see attached photo on how to arrive at the answer.
2nd one Explanation:
4/2 Be 210-206=4 and 84-82=2
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5 Explanation: it is very simple
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Here we have to complete the given radioactive decay reaction: ²³⁸₉₂U → ⁴₂He + ²³⁴₉₀Th → ²³⁴₉₁Pa ? ⁴₂He + ₁¹P + ¹₀n + ⁰₁e + ₋₁⁰e. The completed radioactive reaction is ²³⁸₉₂U → ⁴₂He + ²³⁴₉₀Th → ²³⁴₉₁Pa + ⁴₂He + ₋₁⁰e (electron). In any radioactive reaction the mass number and the atomic number of the reactant and product will be same respectively. The mass number and atomic number of the reactant uranium (U) is 238 and 92 respectively. Thus in the last reaction if ⁴₂He is eliminated then the mass number of the product 234 + 4 – 0 = 238. The atomic number will be 91 + 2 – 1 = 92. Thus the missing particle is electron (₋₁⁰e)