Using nuclear notation, write the balanced nuclear equation representing the alpha decay having the product radon-218.
I have a science final in 2 days and I'm reviewing concepts I don't fully understand. Any help would be great.
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In alpha decay, an radioactive atom will eject an alpha particle, composed of 2 protons and 2 neutrons. That ejected particle, of course, is exactly the same thing as a nucleus of a helium atom. The original atom, in losing 2 protons, transforms into a new element, moving 2 steps to the left on the Periodic Table.
In showing an equation of alpha decay, all the numbers must sum up, as the mass of atoms and particles remains constant. The superscript number of the element is the total number of neutrons and protons, the subscript the total number of protons. So in the equation
222/88 Ra ---> 4/2 He + 218/86 Rn
Radium-222 decays into an alpha (helium) particle and radon-218. Notice that the 2 protons of the alpha particle and the 86 protons of the radon atom add up to 88, the atomic number of radium. Notice also that the total number of neutrons and protons (4 for the alpha particle, 218 for the radon atom) add up to 222, the total number of neutrons and protons for this version of radium.
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