Explanation The first step is alpha decay of U 92 238 . It is shown by the nuclear equation given below. U 92 238 → α T 90 234 h + 2 4 He In the alpha decay, mass number of the nuclide decreases by 4 units and atomic number decreases by 2 units. A helium nucleus is emitted during the alpha decay. The mass number of the daughter nuclide will be 238 − 4 = 234 . The atomic number of the daughter nuclide will be 92 − 2 = 90 . According to the periodic table, the atomic number of Th is 90 . Therefore, the daughter nuclide is T 90 234 h . The second step is beta decay of T 90 234 h . It is shown by the nuclear equation given below. T 90 234 h → 91 234 Pa + − 1 0 e In beta decay, mass number of the daughter nuclide remains same. However atomic number increases by 1 unit. Therefore, the atomic number of daughter nuclide will be 90 + 1 = 91 . According to periodic table, the element with atomic number 91 is Pa . Hence, the daughter nuclide is 91 234 Pa . The third step is beta decay of 91 234 Pa . It is shown by the nuclear equation given below. 91 234 Pa → U 92 234 + − 1 0 e In beta decay, mass number of the daughter nuclide remains same. However atomic number increases by 1 unit. Therefore, the atomic number of daughter nuclide will be 91 + 1 = 92 . According to periodic table, the element with atomic number 92 is uranium. Hence, the daughter nuclide is U 92 234 . The fourth step is alpha decay of U 92 234 . It is shown by the nuclear equation given below. U 92 234 → α T 90 230 h + 2 4 He In the alpha decay, mass number of the nuclide decreases by 4 units and atomic number decreases by 2 units. A helium nucleus is emitted during the alpha decay. The mass number of the daughter nuclide will be 234 − 4 = 230 . The atomic number of the daughter nuclide will be 92 − 2 = 90 . According to the periodic table, the atomic number of Th is 90 . Therefore, the daughter nuclide is T 90 230 h . The fifth step is alpha decay of T 90 230 h . It is shown by the nuclear equation given below. T 90 230 h → α R 88 226 a + 2 4 He In the alpha decay, mass number of the nuclide decreases by 4 units and atomic number decreases by 2 units. A helium nucleus is emitted during the alpha decay. The mass number of the daughter nuclide will be 230 − 4 = 226 . The atomic number of the daughter nuclide will be 90 − 2 = 88 . According to the periodic table, the atomic number of Ra is 88 . Therefore, the daughter nuclide is R 88 226 a . The sixth step is alpha decay of R 88 226 a . It is shown by the nuclear equation given below. R 88 226 a → α R 86 222 n + 2 4 He In the alpha decay, mass number of the nuclide decreases by 4 units and atomic number decreases by 2 units. A helium nucleus is emitted during the alpha decay. The mass number of the daughter nuclide will be 226 − 4 = 222 . The atomic number of the daughter nuclide will be 88 − 2 = 86 . According to the periodic table, the atomic number of Radon is 86 . Therefore, the daughter nuclide is R 86 222 n . The seventh step is alpha decay of R 86 222 n . It is shown by the nuclear equation given below. R 86 222 n → α P 84 218 o + 2 4 He In the alpha decay, mass number of the nuclide decreases by 4 units and atomic number decreases by 2 units. A helium nucleus is emitted during the alpha decay. The mass number of the daughter nuclide will be 222 − 4 = 218 . The atomic number of the daughter nuclide will be 86 − 2 = 84 . According to the periodic table, the atomic number of polonium is 84 . Therefore, the daughter nuclide is P 84 218 o . The eighth step is beta decay of P 84 218 o . It is shown by the nuclear equation given below. P 84 218 o → A 85 218 t + − 1 0 e In beta decay, mass number of the daughter nuclide remains same. However atomic number increases by 1 unit. Therefore, the atomic number of daughter nuclide will be 84 + 1 = 85 . According to periodic table, the element with atomic number 85 is astatine. Hence, the daughter nuclide is A 85 218 t . The ninth step is beta decay of A 85 218 t . It is shown by the nuclear equation given below. A 85 218 t → R 86 218 n + − 1 0 e In beta decay, mass number of the daughter nuclide remains same. However atomic number increases by 1 unit. Therefore, the atomic number of daughter nuclide will be 85 + 1 = 86 . According to periodic table, the element with atomic number 86 is radon. Hence, the daughter nuclide is R 86 218 n

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ISBN: 8220103026918

Author: Walker

Publisher: PEARSON

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Nuclear interactions

In nature, there are certain rules and standards for an interaction. These interactions are universally accepted. There are four types of interactions:

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Chapter 32, Problem 21PCE

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The identification of each of the given 14 decays that occur in the given series.

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Chapter 32, Problem 20PCE

Chapter 32, Problem 22PCE

Chapter 32 Solutions

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Ch. 32.1 - Prob. 1EYU Ch. 32.2 - A given nucleus can decay by alpha decay, beta...Ch. 32.3 - Prob. 3EYU Ch. 32.4 - Prob. 4EYU Ch. 32.5 - Prob. 5EYU Ch. 32.6 - Prob. 6EYU Ch. 32.7 - Prob. 7EYU Ch. 32.8 - Prob. 8EYU Ch. 32.9 - Prob. 9EYU Ch. 32 - Prob. 1CQ

Ch. 32 - Prob. 2CQ Ch. 32 - Prob. 3CQ Ch. 32 - Prob. 4CQ Ch. 32 - Prob. 5CQ Ch. 32 - Prob. 6CQ Ch. 32 - Prob. 7CQ Ch. 32 - Prob. 8CQ Ch. 32 - Prob. 9CQ Ch. 32 - Prob. 1PCE Ch. 32 - Prob. 2PCE Ch. 32 - Prob. 3PCE Ch. 32 - Prob. 4PCE Ch. 32 - Prob. 5PCE Ch. 32 - Prob. 6PCE Ch. 32 - Prob. 7PCE Ch. 32 - Prob. 8PCE Ch. 32 - Prob. 9PCE Ch. 32 - Prob. 10PCE Ch. 32 - Prob. 11PCE Ch. 32 - Prob. 12PCE Ch. 32 - Prob. 13PCE Ch. 32 - Prob. 14PCE Ch. 32 - Prob. 15PCE Ch. 32 - Prob. 16PCE Ch. 32 - Prob. 17PCE Ch. 32 - Prob. 18PCE Ch. 32 - Prob. 19PCE Ch. 32 - Prob. 20PCE Ch. 32 - Prob. 21PCE Ch. 32 - Prob. 22PCE Ch. 32 - Prob. 23PCE Ch. 32 - Prob. 24PCE Ch. 32 - Prob. 25PCE Ch. 32 - Prob. 26PCE Ch. 32 - Prob. 27PCE Ch. 32 - Prob. 28PCE Ch. 32 - Suppose we were to discover that the ratio of...Ch. 32 - A radioactive sample is placed in a closed...Ch. 32 - Radon gas has a half-life of 3.82 d. What is the...Ch. 32 - Prob. 32PCE Ch. 32 - The number of radioactive nuclei in a particular...Ch. 32 - Prob. 34PCE Ch. 32 - Prob. 35PCE Ch. 32 - Prob. 36PCE Ch. 32 - Prob. 37PCE Ch. 32 - Prob. 38PCE Ch. 32 - Prob. 39PCE Ch. 32 - Prob. 40PCE Ch. 32 - Prob. 41PCE Ch. 32 - Prob. 42PCE Ch. 32 - Prob. 43PCE Ch. 32 - Prob. 44PCE Ch. 32 - Prob. 45PCE Ch. 32 - Prob. 46PCE Ch. 32 - Prob. 47PCE Ch. 32 - Prob. 48PCE Ch. 32 - Prob. 49PCE Ch. 32 - Prob. 50PCE Ch. 32 - Prob. 51PCE Ch. 32 - Prob. 52PCE Ch. 32 - Prob. 53PCE Ch. 32 - Prob. 54PCE Ch. 32 - Prob. 55PCE Ch. 32 - Consider a fusion reaction in which two deuterium...Ch. 32 - Prob. 57PCE Ch. 32 - Prob. 58PCE Ch. 32 - Prob. 59PCE Ch. 32 - Prob. 60PCE Ch. 32 - Prob. 61PCE Ch. 32 - Prob. 62PCE Ch. 32 - Prob. 63PCE Ch. 32 - Prob. 64PCE Ch. 32 - Prob. 65PCE Ch. 32 - Prob. 66PCE Ch. 32 - Prob. 67PCE Ch. 32 - Prob. 68GP Ch. 32 - Prob. 69GP Ch. 32 - Prob. 70GP Ch. 32 - Prob. 71GP Ch. 32 - Prob. 72GP Ch. 32 - Prob. 73GP Ch. 32 - Moon Rocks In one of the rocks brought back from...Ch. 32 - Prob. 75GP Ch. 32 - Prob. 76GP Ch. 32 - Prob. 77GP Ch. 32 - Prob. 78GP Ch. 32 - Prob. 79GP Ch. 32 - Prob. 80GP Ch. 32 - Prob. 81GP Ch. 32 - Prob. 82GP Ch. 32 - Prob. 83GP Ch. 32 - Prob. 84GP Ch. 32 - Prob. 85GP Ch. 32 - Prob. 86GP Ch. 32 - Prob. 87GP Ch. 32 - Prob. 88GP Ch. 32 - Prob. 89PP Ch. 32 - Prob. 90PP Ch. 32 - Prob. 91PP

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The identification of each of the given 14 decays that occur in the given series.

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The identification of each of the given 14 decays that occur in the given series.

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Chapter 32 Solutions