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.(89)Ac^(227) is a member of actinium se...

`._(89)Ac^(227)` is a member of actinium series. Another member of the same series of

A

`._(92)U^(235)`

B

`._(90)Th^(232)`

C

`._(89)Ac^(235)`

D

`._(15)P^(34)`

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To solve the problem of identifying another member of the actinium series given that \( _{89}^{227}Ac \) (Actinium-227) is a member, we can follow these steps: ### Step 1: Understand the Actinium Series The actinium series is classified as a 4M + 3 series. This means that when the mass number (A) of an isotope is divided by 4, the remainder should be 3. ### Step 2: Analyze Actinium-227 For Actinium-227, we can check: - Mass number (A) = 227 - When we divide 227 by 4: \[ 227 \div 4 = 56 \quad \text{(quotient)} \quad \text{with a remainder of } 3. \] This confirms that Actinium-227 is indeed part of the 4M + 3 series. ### Step 3: Find Another Member of the Series To find another member of the actinium series, we need to look for another isotope with a mass number that also gives a remainder of 3 when divided by 4. ### Step 4: Check Uranium-235 One well-known isotope that fits this criterion is Uranium-235: - Mass number (A) = 235 - When we divide 235 by 4: \[ 235 \div 4 = 58 \quad \text{(quotient)} \quad \text{with a remainder of } 3. \] This confirms that Uranium-235 is also part of the 4M + 3 series. ### Step 5: Conclusion Thus, another member of the actinium series is \( _{92}^{235}U \) (Uranium-235). ### Final Answer Another member of the actinium series is \( _{92}^{235}U \) (Uranium-235). ---

To solve the problem of identifying another member of the actinium series given that \( _{89}^{227}Ac \) (Actinium-227) is a member, we can follow these steps: ### Step 1: Understand the Actinium Series The actinium series is classified as a 4M + 3 series. This means that when the mass number (A) of an isotope is divided by 4, the remainder should be 3. ### Step 2: Analyze Actinium-227 For Actinium-227, we can check: - Mass number (A) = 227 ...
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