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Question
\[ {}_{86}\mathrm{Rn}^{222} \longrightarrow {}_{\mathrm{y}}\mathrm{Po}^{\mathrm{x}} + {}^{\phantom{+}4}_{+2}\mathrm{He} \]
In the above equation, the values of x and y are:
Options
x = 220, y = 84
x = 218, y = 84
x = 218, y = 82
none of these
MCQ
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Solution
x = 218, y = 84
Explanation:
In alpha decay, the nucleus loses an alpha particle, which is a helium nucleus. So the mass number decreases by 4 and the atomic number decreases by 2. Thus x = 222 − 4 = 218 and y = 86 − 2 = 84.
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