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\ {}_{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:

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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

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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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Chapter 11: Radioactivity - EXERCISE [Page 269]

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 10
Chapter 11 Radioactivity
EXERCISE | Q 24. | Page 269
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