Question

The nucleus \[{}_6{C^{12}}\] absorbs an energetic neutron and emits a beta particle (\[\beta \] ).The resulting nucleus is
A. \[{}_7{N^{14}}\]
B. \[{}_5{B^{13}}\]
C. \[{}_7{N^{13}}\]
D. \[{}_6{C^{13}}\]

Answer 1

Hint: To solve this problem write the equation for the reaction when a beta particle leaves a nucleus. An electron with high speed is called a beta particle when it leaves from the nucleus. The general equation for beta decay is given by, \[_Z{X^A}\; \to {\;_{Z + 1}}X{\prime ^A}\; + {\;_{ - 1}}{e^0}\; + \;{\bar \nu _e}\] where \[A\] and \[Z\] are the mass number and atomic number of the decaying nucleus, and \[X\] and \[X\prime \] are the initial and final elements, respectively.

Complete step by step answer:
We know, in \[{\beta ^ - }\] decay, the weak interaction converts an atomic nucleus into a nucleus with atomic number increased by one, while emitting an electron (\[{e^ - }\]) and an electronic antineutrino (\[\;{\bar \nu _e}\]). \[{\beta ^ - }\] decay generally occurs in neutron-rich nuclei.

Here, we have nucleus \[{}_6{C^{12}}\] which absorbs an energetic neutron and emits a beta particle (\[\beta \] ).So, the reaction equation can be written as, \[{}_6{C^{12}}{ + _0}{n^1} \to {\;_Z}{X^A}\; + {\;_{ - 1}}{e^0}\;\] .Here, let’s assume the daughter nucleus will have the mass number \[A\] and atomic number \[Z\]. Now, we can balance the mass number and atomic number for both sides of the equation. So, balancing mass number we get,
\[12 + 1 = A + 0\]
\[\Rightarrow A = 13\]
Now, balancing the atomic number of the equation from both sides we get,
\[6 + 0 = Z - 1\]
\[\therefore Z = 7\]
So, the resulting nucleus has an atomic number \[7\]. So, it will be a nucleus of nitrogen.Thus, the resulting nucleus will be, \[{}_7{N^{13}}\]

Hence, option C is correct.

Note: \[{\beta ^ - }\] decay occurs when the nucleus has a number of neutrons, greater than protons. Here, the carbon nucleus has an equal number of protons and neutrons, but \[{\beta ^ - }\] occurs due to absorption of extra neutrons. Also, \[{\beta ^ + }\]decay occurs when the nucleus has a number of protons, greater than neutrons. \[{\beta ^ + }\] decay can only happen inside nuclei when the daughter nucleus has a greater binding energy (and therefore a lower total energy) than the mother nucleus.