How do you write the nuclear equations for alpha and beta decay of these isotopes?

Alpha decay emits an alpha particle (A Helium atom):
\({}^4_2mathrm{He}

{}^{14}_6mathrm{C} -> {}^{10}_4mathrm{Be} + {}^4_2mathrm{He}\)
notice that 14 = 4 + 10 and 6 = 2 + 4 (the sides are balanced)
The resultant species is Beryllium-10 because the atomic number is 4.  The mass number is 10, but this only identifies which isotope of Be is present, not the identity of the element.

β- decay emits an electron and an antineutrino (to conserve energy):
\({}^{14}_6mathrm{C} -> {}^{14}_7mathrm{N} + {}^0_{-1}mathrm{{e^{-}}}\) + ¯ve
In this case, the substance decays to Nitrogen-14 (atomic number 7)

β+ decay emits a positron and a neutrino (to conserve energy):
\({}^{14}_6mathrm{C} -> {}^{14}_5mathrm{B} + {}^0_{1}mathrm{{e^{+}}} + {{v}_{e}}\)
In this case, the substance decays to Boron-14 (atomic number 5)

Uranium-248 (\({}^{248}_{92}mathrm{U}\)):
\({alpha}-decay:{}^{248}_{92}mathrm{U} -> {}^{244}_{90}mathrm{Th} + {}^4_2mathrm{He}

{{eta}^{-}} decay: {}^{248}_{92}mathrm{U} -> {}^{248}_{93}mathrm{Np} + {}^0_{-1}mathrm{{e^{-}}} + {^{-}}{{v}_{e}}

{{eta}^{+}} decay: {}^{248}_{92}mathrm{U} -> {}^{248}_{91}mathrm{Pa} + {}^0_{1}mathrm{{e^{+}}} + {{v}_{e}}\)


Potassium-40 (\({}^{40}_{19}mathrm{K}\)):
\({alpha}-decay: {}^{40}_{19}mathrm{K} -> {}^{36}_{17}mathrm{Cl} + {}^4_2mathrm{He}

{{eta}^{-}} decay: {}^{40}_{19}mathrm{K} -> {}^{40}_{20}mathrm{Ca} + {}^0_{-1}mathrm{{e^{-}}} + {^{-}}{{v}_{e}}

{{eta}^{+}} decay: {}^{40}_{19}mathrm{K} -> {}^{40}_{18}mathrm{Ar} + {}^0_{1}mathrm{{e^{+}}} + {{v}_{e}}\)

Post Merge: 10 years ago

Just an aside, when someone refers to beta-decay and doesn't specify +/-, it's assumed they're talking about β- decay (emission of an electron and an antineutrino).

Also note that the neutrino and antineutrinos are only to balance energy/matter change, and do not affect the identity or mass of any of the species of the equation.

Post Merge: 10 years ago

I'm going to lose my mind.  This is a bug in the system.  The backslashes in LaTex are removed if the system does a post merge.  I can't edit my original post (what sense does that make?) and I can't access its source, so all this work looks like a jumbled mess.  This is deeply annoying.

Alpha decay emits an alpha particle (A Helium atom):
\({}^4_2\mathrm{He}

{}^{14}_6\mathrm{C} -> {}^{10}_4\mathrm{Be} + {}^4_2\mathrm{He}\)
notice that 14 = 4 + 10 and 6 = 2 + 4 (the sides are balanced)
The resultant species is Beryllium-10 because the atomic number is 4.  The mass number is 10, but this only identifies which isotope of Be is present, not the identity of the element.

β- decay emits an electron and an antineutrino (to conserve energy):
\({}^{14}_6\mathrm{C} -> {}^{14}_7\mathrm{N} + {}^0_{-1}\mathrm{{e^{-}}}\) + ¯ve
In this case, the substance decays to Nitrogen-14 (atomic number 7)

β+ decay emits a positron and a neutrino (to conserve energy):
\({}^{14}_6\mathrm{C} -> {}^{14}_5\mathrm{B} + {}^0_{1}\mathrm{{e^{+}}} + {{v}_{e}}\)
In this case, the substance decays to Boron-14 (atomic number 5)

Uranium-248 (\({}^{248}_{92}\mathrm{U}\)):
\({alpha}-decay:{}^{248}_{92}\mathrm{U} -> {}^{244}_{90}\mathrm{Th} + {}^4_2\mathrm{He}

{{eta}^{-}} decay: {}^{248}_{92}\mathrm{U} -> {}^{248}_{93}\mathrm{Np} + {}^0_{-1}\mathrm{{e^{-}}} + {^{-}}{{v}_{e}}

{{eta}^{+}} decay: {}^{248}_{92}\mathrm{U} -> {}^{248}_{91}\mathrm{Pa} + {}^0_{1}\mathrm{{e^{+}}} + {{v}_{e}}\)


Potassium-40 (\({}^{40}_{19}\mathrm{K}\)):
\({alpha}-decay: {}^{40}_{19}\mathrm{K} -> {}^{36}_{17}\mathrm{Cl} + {}^4_2\mathrm{He}

{{eta}^{-}} decay: {}^{40}_{19}\mathrm{K} -> {}^{40}_{20}\mathrm{Ca} + {}^0_{-1}\mathrm{{e^{-}}} + {^{-}}{{v}_{e}}

{{eta}^{+}} decay: {}^{40}_{19}\mathrm{K} -> {}^{40}_{18}\mathrm{Ar} + {}^0_{1}\mathrm{{e^{+}}} + {{v}_{e}}\)