Kelvi

In the nuclear emission stated below, the mass number and atomic number of the product Q respectively, are: $\\ ^{200}_{82}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{\beta^-} P \xrightarrow{e^-} Q$

hard

Nuclei

2024

physics

288, 82

286, 81

280, 81

286, 80

Explanation

To solve this problem, we need to analyze the nuclear reactions step-by-step. $\\$ The initial nucleus is $^{200}_{82}X. \\$ 1. Alpha decay: $\\$ In alpha decay, an alpha particle $(^{4}_{2}\alpha)$

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In the nuclear emission stated below, the mass number and atomic number of the product Q respectively, are:82200X→αY→e+Z→β−P→e−Q \\ ^{200}_{82}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{\beta^-} P \xrightarrow{e^-} Q82200​Xα​Ye+​Zβ−​Pe−​Q

Explanation

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In the nuclear emission stated below, the mass number and atomic number of the product Q respectively, are: $\\ ^{200}_{82}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{\beta^-} P \xrightarrow{e^-} Q$