lithium
What is missing is the type of decay that occurs during the transformation. For example, uranium-238 decays into thorium-234 through alpha decay, so the missing component would be the emission of an alpha particle in the balanced equation.
If radon-210 undergoes alpha decay, it will produce the alpha particle (which is a helium-4 nucleus) and polonium-206. The equation looks like this: 86210Ra => 24He + 84206Po You'll note that in the balanced nuclear equation, the atomic numbers, which are the subscripts, balance on both sides of the equation (86 = 2 + 84). The atomic masses, which are the superscripts, also balance on both sides of the equation (210 = 4 + 206).
The other product of the radioactive decay of 204 Pb to 200 Hg is an alpha particle (4 He nucleus). This decay process involves the emission of an alpha particle, leading to the formation of 200 Hg.
224
An alpha particle consists of two protons and two neutrons.
What is missing is the type of decay that occurs during the transformation. For example, uranium-238 decays into thorium-234 through alpha decay, so the missing component would be the emission of an alpha particle in the balanced equation.
The decay is:Bi-187------------------Tl-183
If radon-210 undergoes alpha decay, it will produce the alpha particle (which is a helium-4 nucleus) and polonium-206. The equation looks like this: 86210Ra => 24He + 84206Po You'll note that in the balanced nuclear equation, the atomic numbers, which are the subscripts, balance on both sides of the equation (86 = 2 + 84). The atomic masses, which are the superscripts, also balance on both sides of the equation (210 = 4 + 206).
The other product of the radioactive decay of 204 Pb to 200 Hg is an alpha particle (4 He nucleus). This decay process involves the emission of an alpha particle, leading to the formation of 200 Hg.
92U238 --------> 90Th234 + 2He4
The balanced equation for the alpha decay of thorium-229, Th-229, is: Th-229 -> Ra-225 + He-4 This equation shows that a thorium-229 nucleus undergoes alpha decay to form a radium-225 nucleus and a helium-4 particle.
To balance the nuclear equation, a beta particle (negatron) must be included. The balanced equation would be 220/88 Ra -> 4/2 He (alpha particle) + 212/86 Rn + 2 -1 e.
alpha particle
alpha particle
The equation for the beta decay of 137Cs:55137Cs --> 56137Ba + -10e where the e is a negative beta particle or electron.
An alpha particle has two protons.
There are a number of radioactive isotopes of copper, choosing 66Cu as on that undergoes negative beta decay, the equation is: 2966Cu --> 3066Zn + -10e Where e represents the beta particle, which can also be viewed as an electron.