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Which of the following represents alpha decay?

A. [tex]_{43}^{90} Tc \rightarrow _{43}^{90} Tc + \gamma[/tex]

B. [tex]_{64}^{160} Gd \rightarrow _{65}^{160} Tb + ^0 e[/tex]

C. [tex]_{63}^{150} Eu + _{-1}^0 e \rightarrow _{62}^{150} Sm[/tex]

D. [tex]_{64}^{148} Gd \rightarrow _{62}^{144} Sm + _2^4 He[/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's break down the options to determine which one represents alpha decay.

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, which is equivalent to a helium nucleus represented as [tex]\( {}_2^4 He \)[/tex].

Let's analyze each option:

Option A: [tex]\( {}_{43}^{90} Tc \rightarrow {}_{43}^{90} Tc + \gamma \)[/tex]
- Here, Technetium-90 ([tex]\( {}_{43}^{90} Tc \)[/tex]) is decaying and emitting a gamma photon ([tex]\( \gamma \)[/tex]).
- This is an example of gamma decay, not alpha decay because it involves the emission of a gamma photon, not an alpha particle.

Option B: [tex]\( {}_{64}^{160} Gd \rightarrow {}_{65}^{160} Tb + {}^0 e \)[/tex]
- Here, Gadolinium-160 ([tex]\( {}_{64}^{160} Gd \)[/tex]) is decaying into Terbium-160 ([tex]\( {}_{65}^{160} Tb \)[/tex]) and emitting an electron ([tex]\( {}^0 e \)[/tex]).
- This is an example of beta decay, specifically beta-plus decay or positron emission because the nucleus emits a beta particle (an electron), and the atomic number increases by 1.

Option C: [tex]\( {}_{63}^{150} Eu + {}_{-1}^0 e \rightarrow {}_{62}^{150} Sm \)[/tex]
- Here, Europium-150 ([tex]\( {}_{63}^{150} Eu \)[/tex]) captures an electron ([tex]\( {}_{-1}^0 e \)[/tex]) and converts into Samarium-150 ([tex]\( {}_{62}^{150} Sm \)[/tex]).
- This is an example of electron capture or possibly positron emission as it involves an electron, not the emission of an alpha particle.

Option D: [tex]\( {}_{64}^{148} Gd \rightarrow {}_{62}^{144} Sm + {}_2^4 He \)[/tex]
- Here, Gadolinium-148 ([tex]\( {}_{64}^{148} Gd \)[/tex]) is decaying into Samarium-144 ([tex]\( {}_{62}^{144} Sm \)[/tex]) and emitting a helium nucleus ([tex]\( {}_2^4 He \)[/tex]).
- This is a clear representation of alpha decay because the nucleus emits an alpha particle, consisting of 2 protons and 2 neutrons.

Thus, the correct representation of alpha decay is Option D:
[tex]\[ {}_{64}^{148} Gd \rightarrow {}_{62}^{144} Sm + {}_2^4 He \][/tex]
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