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15 July, 10:29

Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β particles are produced in the complete decay series? α particles β particles?

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  1. 15 July, 11:45
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    8 alpha particles

    4 beta particles

    Explanation:

    We are given;

    Neptunium-237 Thallium-205 Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

    We are required to determine the number of beta and alpha particles produced to complete the decay series.

    We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2. When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

    In this case;

    Neptunium-237 has an atomic number 93, while,

    Thallium-205 has an atomic number 81.

    Therefore;

    ²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

    We can get x and y

    237 = 4x + y (0) + 205

    237-205 = 4x

    4x = 32

    x = 8

    On the other hand;

    93 = 2x + (-y) + 81

    but x = 8

    93 = 16 - y + 81

    y = 4

    Therefore, the complete decay equation is;

    ²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

    Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.
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