QUESTION IMAGE
Question
question 5 (3 points)
fill in all of the missing information; 2 symbols and 2 numbers (from the letter choices below)
210 206 b 0
rn → po + c + d
86 a 2 0
a = 1
b = 2
c = 3
d = 4
a. 0 b. 1 c. -1 d. 2 e. 3 f. 4 g. beta
h. gamma i. alpha j. h k. rn l. np m. po n. cm
o. 82 p. 83 q. 84 r. 92 s. 206 t. 208 u. 210
v. 212 w. 416
Step1: Balance atomic numbers
In a nuclear - decay reaction, the sum of atomic numbers on the left - hand side equals the sum of atomic numbers on the right - hand side. The atomic number of Rn is 86. For the product side, if we assume the atomic number of Po is \(x\), and we know the other particles contribute to the atomic number. Since Rn (\(Z = 86\)) decays, and one of the products has an atomic number of 2. Let the atomic number of Po be \(x\). Then \(86=x + 2+0\), so \(x = 84\). The atomic number of Po is 84, so \(A = 84\).
Step2: Balance mass numbers
The mass number of Rn is 210. The mass number of Po is 206. Let the mass number of the particle \(C\) be \(y\) and of \(D\) be \(z\). Then \(210=206 + y+z\). Since one of the particles has a mass number of 4 (an alpha - particle) and the other has a mass number of 0 (a gamma - ray), the particle \(C\) with mass number 4 and atomic number 2 is an alpha - particle (\(\alpha\)), and \(D\) with mass number 0 and atomic number 0 is a gamma - ray (\(\gamma\)).
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- Q. 84
- S. 206
- I. alpha
- H. gamma