15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T65e,

To determine the number of neutrons in the atom after the emission of a $$\beta$$-particle and an $$\alpha$$-particle from $${ }_{83}^{214} \mathrm{Bi}$$, we need to understand how these emissions affect the atomic and mass numbers.

1. Emission of a $$\beta$$-particle:

A $$\beta$$-particle is essentially an electron, and its emission converts a neutron into a proton. Therefore, the atomic number increases by 1, but the mass number remains the same. For $${ }_{83}^{214} \mathrm{Bi}$$ (bismuth), after the emission of a $$\beta$$-particle, the atomic number becomes 84:

$$ { }_{83}^{214} \mathrm{Bi} \rightarrow { }_{84}^{214} \mathrm{Po} + \beta $$

2. Emission of an $$\alpha$$-particle:

An $$\alpha$$-particle consists of 2 protons and 2 neutrons, so its emission decreases the atomic number by 2 and the mass number by 4. Therefore, for $${ }_{84}^{214} \mathrm{Po}$$ (polonium), after the emission of an $$\alpha$$-particle, the atomic number becomes 82 and the mass number becomes 210:

$$ { }_{84}^{214} \mathrm{Po} \rightarrow { }_{82}^{210} \mathrm{Pb} + \alpha $$

Now, we need to find the number of neutrons in the resulting atom, $${ }_{82}^{210} \mathrm{Pb}$$ (lead). The number of neutrons is given by the difference between the mass number and the atomic number:

$$ \text{Number of neutrons} = \text{Mass number} - \text{Atomic number} $$

$$ \text{Number of neutrons} = 210 - 82 = 128 $

Radioactivity And Nuclear Chemistry

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