Here we go:
when the following nuclear reaction is completed, the missing element will be:
___+ 4^He ---> 2 v0 1^n+ 244^Cf
Okay, so im given the choices of 242^Cm, 242^Am, 240^Es or 242^Bk....
I already determined that it's not 240 Es (or at least i think I did...) could someone please explain to me what the right answer is? As two of the choices have the same atomic weight and im just not sure how to choose..
These are almost impossible to type on this board so I'll not try. I've never been very successful at it. However, here is what you do.
The atomic numbers (the numbers at the lower left) on the left side of the equation must add up to the same number on the right. The mass numbers (the numbers at the upper left) must add up on left and right side of the equation.
Let's do it all atomic number first.
______ + 2He ==> 2 0n + 98Cf (you must look on the periodic table and see that He is element #2 and Cf is element #98). Now we add them up. Since the unknown is on the left, we add on the right to see what we must have on the left. There is 2*0 + 98 = 98 on the right. So on the left 2 + something must be 98. Obviously the something is 96. So what is element 96 on the periodic table. I believe that is Cm.
Now for the top.
Cm + 4He ==> 2 1n + 244Cf
So on the right we have 2*1 + 244 = 246. On the left we have 4 + something = 246 and the something obviously must be 242. So the full equation looks this way if you let me place the atomic numbers on the left and mass numbers on the right. Check it to see that it all adds up.
96Cm^242 + 2He^4 ==> 2 0n^1 + 98Cf^244.
You see it's as simple as adding the bottom row and the top rows and making each come out the same on both sides of the equation. Let me know if there is something about this you don't understand.
ohh, no I totally get it now. I was just looking at the atomic weight and i didn't even think about the atomic numbers XD thank u so much! this was very helpful!
To determine the missing element in the nuclear reaction, we need to understand the pattern of nucleon numbers (mass numbers) and atomic numbers. Let's break it down step by step.
The given nuclear reaction is:
___ + 4^He ---> 2 v0 1^n + 244^Cf
Here, the symbol "___" represents the missing element that we need to identify.
Let's look at the given choices:
1. 242^Cm (Curium) - Atomic number: 96, Mass number: 242
2. 242^Am (Americium) - Atomic number: 95, Mass number: 242
3. 240^Es (Einsteinium) - Atomic number: 99, Mass number: 240
4. 242^Bk (Berkelium) - Atomic number: 97, Mass number: 242
Now, let's analyze the information from the nuclear reaction:
In the reactants, we have 4^He and the missing element, represented by "___". 4^He is a helium nucleus, which means it consists of 2 protons (atomic number) and 2 neutrons (mass number).
In the products, we have 2 v0 1^n (neutrons) and 244^Cf (Californium).
Since neutrons do not have any protons, we know that the sum of protons in the reactants should equal the sum of protons in the products.
The mass number (sum of protons and neutrons) should also be conserved in the reaction.
In the reactants, the atomic number of the missing element + 2 (from 4^He) should be equal to the sum of atomic numbers in the products (2 from v0 1^n and some atomic number from 244^Cf).
Looking at the choices:
1. 242^Cm: If the missing element is 242^Cm, the sum of atomic numbers would be 96 + 2 = 98, which is not equal to 2 + atomic number of 244^Cf.
2. 242^Am: If the missing element is 242^Am, the sum of atomic numbers would be 95 + 2 = 97, which matches the atomic number of 242^Bk, one of the product elements. It could be a possible answer.
3. 240^Es: We can discard this choice since its mass number does not match the 244^Cf in the products.
4. 242^Bk: We already used this atomic number (97) to check against 242^Am.
Therefore, based on the above analysis, the correct answer is 242^Am (Americium). The missing element is 242^Am, and the complete reaction becomes:
242^Am + 4^He ---> 2 v0 1^n + 244^Cf