Math

QuestionQUESTION { }^{\circ} ANSWER
In the process of nuclear or radioactive decay, an unstable nucleus spontaneously emits a particle. When this occurs, the nucleus of one element can change into the nucleus of a different element. The resulting change of one element to another is called transmutation. The nuclear decay can be represented by a nuclear equation using atomic symbols for the nuclei and the emitted particles.
Radioactive decay: radioactive nucleus \rightarrow new nucleus + emitted particle
In a nuclear equation, the sum of the mass numbers and the sum of the atomic numbers on one side of the equation must equal the sum of the mass numbers and the sum of the atomic numbers, respectively, on the other side of the equation. 1985 K10e+2038Ca{ }_{19}^{85} \mathrm{~K} \rightarrow-{ }_{-1}^{0} e+{ }_{20}^{38} \mathrm{Ca} 103 K10e+B3Ar{ }_{10}^{3} \mathrm{~K} \rightarrow{ }_{1}^{0} e+{ }_{B}^{3} \mathrm{Ar} 183 K{4He+174Cl{ }_{18}^{3} \mathrm{~K} \rightarrow{ }_{\{ }^{4} \mathrm{He}+{ }_{17}^{4} \mathrm{Cl} 1938 K+10e2138Ca{ }_{19}^{38} \mathrm{~K}+{ }_{1}^{0} e \rightarrow{ }_{21}^{38} \mathrm{Ca} IDON'T KNOWYET
Identify the balanced nuclear equation for the positron emission of 38K{ }^{38} \mathbf{K}.

Studdy Solution
The equation 1938K+10e+1838Ar _{19}^{38} \mathrm{K} \rightarrow _{+1}^{0} e + _{18}^{38} \mathrm{Ar} represents the positron emission of 38K ^{38} \mathrm{K} .
The balanced nuclear equation for the positron emission of 38K ^{38} \mathrm{K} is:
1938K+10e+1838Ar \boxed{_{19}^{38} \mathrm{K} \rightarrow _{+1}^{0} e + _{18}^{38} \mathrm{Ar}}

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