When Uranium 238 decays, it emits 8 alpha particles and 6 beta particles before finally becoming the stable daughter product; What is the atomic number and mass number of the daughter product? Show your work.
92U238 ==> 2He4 + 90X234
90X234 ==> 2He4 + 88Y230
88Y230 ==> 2He4 + 86Z226
etc. Subtract 2 for atomic number and 4 for mass number.
For beta particle, you increase atomic number by 1 and leave mass number the same. Follow the series down the chain until you get to the end with 8 alpha and 6 beta emissions.
OR you can do it this way.
For alpha. 8 x -2 protons = -16 protons
................8 x -2 neutrons = -16 neutrons or -32 for mass number.
For beta. 6 x +1 proton = 6 protons.
So 92U238
238-32 = 206 for mass number.
92 -16 + 6 = 82 for the atomic number and you're left with 82X206. What's X? Look on the periodic chart and find atomic number of 82 is Pb so the chain ends at 82Pb206
each α-particle is composed of two protons and two neutrons
... a helium nucleus
a beta particle is an electron from the decay of a neutron
... this results in a proton and the beta electron
32 nucleons are emitted ... 16 protons and 16 neutrons
6 of the remaining neutrons decay into protons
To determine the atomic number and mass number of the daughter product, we need to consider the decay process of Uranium-238.
Uranium-238 (symbol U-238) undergoes alpha decay, where it emits an alpha particle (two protons and two neutrons). This results in the formation of a daughter product. Given that Uranium-238 emits 8 alpha particles during its decay process, it means that 8 alpha decays have occurred.
One alpha decay reduces the atomic number by 2 and the mass number by 4. Therefore, 8 alpha decays will decrease the atomic number by 8 x 2 = 16 and the mass number by 8 x 4 = 32.
Now, let's calculate the atomic number and mass number of the daughter product:
Atomic number of the daughter product = Atomic number of Uranium-238 - (Number of alpha decays x 2)
Mass number of the daughter product = Mass number of Uranium-238 - (Number of alpha decays x 4)
Atomic number of Uranium-238 = 92
Mass number of Uranium-238 = 238
Atomic number of the daughter product = 92 - (8 x 2) = 92 - 16 = 76
Mass number of the daughter product = 238 - (8 x 4) = 238 - 32 = 206
Therefore, the atomic number of the daughter product is 76, and the mass number is 206.
To determine the atomic number and mass number of the daughter product when Uranium 238 decays, we need to know the atomic number and mass number of Uranium 238 itself.
Uranium 238 has:
- Atomic number (number of protons) = 92
- Mass number (number of protons + neutrons) = 238
Alpha decay involves the emission of an alpha particle, which consists of 2 protons and 2 neutrons. Therefore, each alpha particle emitted during the decay reduces the atomic number of the parent element by 2 and the mass number by 4.
In this case, Uranium 238 emits 8 alpha particles. So,
- Atomic number of the daughter product = Atomic number of Uranium 238 - (8 * 2) = 92 - 16 = 76
- Mass number of the daughter product = Mass number of Uranium 238 - (8 * 4) = 238 - 32 = 206
After alpha decay, the daughter product has an atomic number of 76 and a mass number of 206.
Now, let's consider the beta decay. Beta decay involves the emission of a beta particle, which is an electron or a positron. It doesn't affect the mass number of the parent element, but it increases the atomic number by 1.
In this case, Uranium 238 emits 6 beta particles. So,
- Atomic number of the daughter product = Atomic number of the previous daughter product + (6 * 1) = 76 + 6 = 82 (Note that the atomic number was reduced to 76 after alpha decay.)
Since the atomic number of 82 corresponds to the chemical element Lead (Pb), the stable daughter product of Uranium 238 decay is Lead (Pb).
Therefore, the atomic number of the daughter product is 82 and the mass number is 206, as determined previously.