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\({ }_{11} \mathrm{Na}^{24} \longrightarrow{ }_{\mathrm{Q}} \mathrm{Mg}^{\mathrm{P}}+{ }_{-1} \mathrm{e}^{0}\) In the above equation, the values of P and Q are:

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Question

\({ }_{11} \mathrm{Na}^{24} \longrightarrow{ }_{\mathrm{Q}} \mathrm{Mg}^{\mathrm{P}}+{ }_{-1} \mathrm{e}^{0}\)

In the above equation, the values of P and Q are:

Options

  • P = 25, Q = 12

  • P = 24, Q = 10

  • P = 24, Q = 12

  • none of these

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Solution

P = 24, Q = 12

Explanation:

In beta decay, a neutron changes into a proton and an electron. The proton stays in the nucleus, so the mass number remains 24, but the atomic number increases by 1, giving Q = 11 + 1 = 12.

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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 23. | Page 269
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