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241 Am decays by alpha particle emission. The daughter isotope in this case is (the atomic numbers of Np, Pu, Am, Cm, and Bk are 91, 92, 93, 94, and 95 respectively) O 235 Bk. O 245 Bk. O 239Np. O not given. O 237Np.

241 Am decays by alpha particle emission. The daughter isotope in this case is (the atomic numbers of Np, Pu, Am, Cm, and Bk are 91, 92, 93, 94, and 95 respectively) O 235 Bk. O 245 Bk. O 239Np. O not given. O 237Np.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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**Alpha Decay of Americium-241 (241^Am)** In the process of alpha decay, an Americium-241 (^241Am) atom decays by emitting an alpha particle. As a result, it transforms into a different element. The atomic numbers of pertinent elements are given as follows: Neptunium (Np) - 91, Plutonium (Pu) - 92, Americium (Am) - 93, Curium (Cm) - 94, and Berkelium (Bk) - 95. Question: **Identify the daughter isotope in this case from the following options:** 1. \( \circ \) \( ^{235}\text{Bk} \) 2. \( \circ \) \( ^{245}\text{Bk} \) 3. \( \circ \) \( ^{239}\text{Np} \) 4. \( \circ \) not given 5. \( \circ \) \( ^{237}\text{Np} \) With the emission of an alpha particle (which consists of 2 protons and 2 neutrons), the atomic number of the new element decreases by 2 and the mass number decreases by 4 compared to the original element. For ^241Am (atomic number 95): - The atomic number after decay: \( 95 - 2 = 93 \) (which corresponds to Neptunium). - The mass number after decay: \( 241 - 4 = 237 \). Thus, the resulting isotope is \( ^{237}\text{Np} \). Correct Answer: \( \circ \) \( ^{237}\text{Np} \)
Transcribed Image Text:**Alpha Decay of Americium-241 (241^Am)** In the process of alpha decay, an Americium-241 (^241Am) atom decays by emitting an alpha particle. As a result, it transforms into a different element. The atomic numbers of pertinent elements are given as follows: Neptunium (Np) - 91, Plutonium (Pu) - 92, Americium (Am) - 93, Curium (Cm) - 94, and Berkelium (Bk) - 95. Question: **Identify the daughter isotope in this case from the following options:** 1. \( \circ \) \( ^{235}\text{Bk} \) 2. \( \circ \) \( ^{245}\text{Bk} \) 3. \( \circ \) \( ^{239}\text{Np} \) 4. \( \circ \) not given 5. \( \circ \) \( ^{237}\text{Np} \) With the emission of an alpha particle (which consists of 2 protons and 2 neutrons), the atomic number of the new element decreases by 2 and the mass number decreases by 4 compared to the original element. For ^241Am (atomic number 95): - The atomic number after decay: \( 95 - 2 = 93 \) (which corresponds to Neptunium). - The mass number after decay: \( 241 - 4 = 237 \). Thus, the resulting isotope is \( ^{237}\text{Np} \). Correct Answer: \( \circ \) \( ^{237}\text{Np} \)
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