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29080⁢X -> [š›¼] Y -> [e+] -> Z -> [š›½āˆ’] -> P -> [eāˆ’] -> Q In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are:

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Question

\[\ce{^290_80X ->[\alpha] Y ->[e+] Z ->[\beta-] P ->[e-] Q}\]

In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are:

Options

  • 288, 82

  • 286, 81

  • 280, 81

  • 286, 80

MCQ
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Solution

286, 81

Explanation:

\[\ce{^290_80X ->[\alpha] ^286_80Y ->[e+] ^286_79Z ->[\beta-] ^286_80P ->[e-] ^286_81Q}\]

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