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3 years ago

When 38^88Sr decays to 34^84Kr, a(n) ______________ is emitted

Chemistry

1 answer:

navik [9.2K]3 years ago

3 0

Answer:

The correct answer is - alpha particle and positron.

Explanation:

In this question, it is given that, 38^88Sr decays to 34^84Kr, which means there is an atomic number decrease by 4, 38 to 34, and atomic mass decreases by 4 as well 88 to 84.

A decrease in the atomic mass is possible only when there is an emission of the alpha particle as an alpha particle is made of 2 protons and 2 neutrons. If an atom emits an alpha particle, there is a change in atomic number as it decreases by two, and its mass number decreases by four.

So after the emission of an alpha particle, the new atom would be

38^88Sr=> 36^84X => 34^84Kr

so there is also two positron emission that leads to decrease in atomic number by one with each emission:

38^88Sr=> 2^4He+ 36^84X => 36^84X + 2(1^0β+) => 34^84Kr

Positron decay is the conversion of a proton into a neutron with the emission of a positron that causes the atomic number is decreased by one, which causes a change in the elemental identity of the daughter isotope.

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Explanation:

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$-r_A=k\times\left[A\right]^\alpha\times\left[B\right]^\beta\bigm$

Replacing for the data

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Expression 2 $0.01280=k\times{0.24}^\alpha\times{0.22}^\beta$

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