Write a nuclear equation for the indicated decay of each of the following nuclides.
Th−234 (beta)......no clue
O−15 (positron emission): 16/8 O-->16/7N + B+ (said was wrong)
Ar-37: 37/18Ar + e ---> 37/17Cl + v (said was wrong)
what would be the proper equations?
Following my post above.
atomic number on left, element in middle, mass number on right. Left numbers must add up on both side; right numbers must add up on both sides. So for Th we write 90Th234 ==> beta + ?
beta is an electron which is -1e0.
90Th234 ==> -1e0 + AXZ
So 90 = -1 + A and A must be 90 + 1 = 91
234 = 0 + Z so Z must be 234
The element which has atomic number 91 is Pa so the full equation is
90Th234 ==> -1e0 + 91Pa234.
The others are done the same way. I'll be glad to check your answers.
Th-234 (beta) decay:
Th-234 undergoes beta decay, where a neutron is converted into a proton, and an electron and an electron antineutrino are emitted.
The nuclear equation is:
^234/90Th → ^234/91Pa + 0/-1β
O-15 (positron emission):
O-15 undergoes positron emission, where a proton is converted into a neutron, and a positron and an electron neutrino are emitted.
The nuclear equation is:
^15/8O → ^15/7N + 0/+1e + 0/0ν
Ar-37:
Ar-37 undergoes electron capture, where an electron is captured by the nucleus, causing a proton to convert into a neutron, and an electron neutrino is emitted.
The nuclear equation is:
^37/18Ar + 0/-1e → ^37/17Cl + 0/0ν
Please note that the charges and mass numbers have been written in the superscript and subscript respectively.
To write a correct nuclear equation for the indicated decay of each nuclide, you need to understand the type of decay and the resulting daughter nucleus. Let's go through each case:
1. Th-234 (beta decay): Beta decay is the emission of a beta particle, which is an electron (e-) or a positron (e+). In the case of Th-234, it undergoes beta decay and emits an electron. The equation would be:
234/90 Th → 234/91 Pa + 0/-1 e
2. O-15 (positron emission): Positron emission involves the emission of a positron (e+). However, in your initial attempt, you wrote the equation incorrectly. The correct equation is:
15/8 O → 15/7N + 0/+1 e
3. Ar-37 (electron capture): Electron capture is when an electron is captured by the nucleus. In this case, Ar-37 undergoes electron capture and the equation would be:
37/18 Ar + 0/-1 e → 37/17Cl + v
Note: The "v" symbol represents a neutrino, which is also produced in some decay processes.
When writing nuclear equations, it's important to balance both mass number (the sum of protons and neutrons) and atomic number (the number of protons) on both sides of the equation to ensure conservation of mass and charge.