Question:
In a radioactive decay chain, $^{232}_{90}Th$ nucleus decays to $^{212}_{92}Pb$ nucleus. Let $N_{\alpha}$ and $N_{\beta}$ be the number of $alpha$ and $\beta^{-}$ particles, respectively, emitted in this decay process. Which of the following statements is (are) true?
In a radioactive decay chain, $^{232}_{90}Th$ nucleus decays to $^{212}_{92}Pb$ nucleus. Let $N_{\alpha}$ and $N_{\beta}$ be the number of $alpha$ and $\beta^{-}$ particles, respectively, emitted in this decay process. Which of the following statements is (are) true?
Updated On: Jun 14, 2022
- $N_{\alpha} = 5 $
- $N_{\alpha} = 6 $
- $N_{\beta} = 2$
- $N_{\beta} = 4$
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The Correct Option is C
Solution and Explanation
$^{232}_{90}Th$ Th is converting into $^{212}_{82}Pb$
Change in mass number (A) = 20
$\therefore$ no of $\alpha $ particle $ = \frac{20}{4} = 5 $
Due to $5 \, \alpha$ particle, z will change by 10 unit.
Since given change is 8, therefore no. of $\beta$ particle is 2
Change in mass number (A) = 20
$\therefore$ no of $\alpha $ particle $ = \frac{20}{4} = 5 $
Due to $5 \, \alpha$ particle, z will change by 10 unit.
Since given change is 8, therefore no. of $\beta$ particle is 2
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In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
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Following are the terms related to nucleus:
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