I need help completing nuclear equations:
a. ? + 59/27Co -> 56/25Mn + 4/2He
b. ? -> 14/7N + 0/-1e
(I think this one is 14/6C, but I'm not sure.)
c. 0/-1e + 76/36Kr -> ?
d. 4/2He + 241/95Am -> ? + 2 1/0 n
These are easier than they look. The upper numbers must add up on both sides and the lower numbers must add up on both sides.
a. ? + 59/27Co -> 56/25Mn + 4/2He
upper numbers: On the right you have 56 + 4 = 60 so the left must be 59 + ? = 60. ? must be 1.
lower numbers: On the right 25 + 2 = 27 so th left must be 27 + ? = 27. ? must be 0. What element is 27? Must be Co. So what has a mass of 1 and a charge of 0? must be a neutron.
1/0n + 59/27Co -> 56/25Mn + 4/2He Things can't get much easier than that. Your solution to b is correct. I'll leave the others for you. Shout if you have trouble.
Thank you so much for your help!
So would C be 76/37Rb?
I'm still having trouble with d though. 4/2He would make 241/27Co into 237/25 Mn, right? But I'm confused by the 2 1/0 n
I think you're confused by more than the 2 1/0 n. Where did the Co come from? That isn't in the problem. Co is in another problem. Where did the Mn come from? That Am is americium and you've ignored that.
"I'm still having trouble with d though. 4/2He would make 241/27Co into 237/25 Mn, right? But I'm confused by the 2 1/0 n"
Here is what I would do.
4/2He + 241/95Am -> ? + 2 1/0 n .
If that 2 1/0 n is confusing you let's write it this way.
4/2He + 241/95Am -> ? + 1/0 n + 1/0 n
upper numbers: on the left we have 4 + 241 = 245. On the right we have ? + 1 + 1 so ? must be 243.
lower numbers: on the left we have 2 + 95 = 97. On the right we have ? + 0 + 0 so ? must be 97. What's atomic number 97 on the periodic table. That's Bk, Berkelium or 243/97 Bk. Final looks this way.
4/2He + 241/95Am -> 243/97Bk + 2 1/0 n
No, c isn't correct. Here is what you have.
0/-1e + 76/36Kr -> 76/37Rb but you didn''t add correctly. -1 + 36 = 35 and not 37 so it must be Br and not Rb
Ah, thank you so much! I didn't realize I mixed up problems c and d when I was writing the question.
For c, I thought that for beta decay, we would add 1 to the atomic number.. if it's on the left side of the equation, I should subtract 1?
For beta DECAY you DO add one because you're keeping the total on the left = to total on the right as in this equation.
238/92 U ==> 238/93 Np + 0/-1e but when you're ADDING a beta particle to something on the left such as
0/-1e + 76/36Kr ==> 76/35Br you must START with a higher number to account for the negative charge on the electrons. Your rule of thumb is good for DECAY and it make it easy to pronounce quickly what the product is but just remember it's reversed when you reverse the reaction. To be safe always remember the numbers on both sides must add up. That gets it right EVERY time.
Let's go through each equation one by one and determine the missing elements.
a. ? + 59/27Co -> 56/25Mn + 4/2He
To balance this equation, the total atomic numbers and the total mass numbers of both sides should be equal.
Starting with the left side, we have an unknown element (+ ?) and Cobalt-59 (^59/27Co). The atomic number of Cobalt is 27, so our unknown element should have an atomic number of (27 + unknown).
Next, we have Manganese-56 (^56/25Mn) and Helium-4 (^4/2He) on the right side. The atomic number of Manganese is 25, so we can conclude that the unknown element should have an atomic number of (25 + unknown).
Now let's balance the mass numbers. The mass number of Cobalt is 59, so the mass number of our unknown element plus 59 should be equal to the mass number of Manganese (56). In other words, (mass number of unknown + 59) = 56. Simplifying this equation, we get the mass number of the unknown element as (56 - 59 = -3).
Based on this information, the unknown element is an element with an atomic number of (27 + unknown) and a mass number of (-3).
b. ? -> 14/7N + 0/-1e
In this equation, we know that Nitrogen-14 (^14/7N) is formed and an electron (^0/-1e) is emitted.
The atomic number of Nitrogen-14 is 7, so the unknown element should have an atomic number that is one less than 7. This means the atomic number of the unknown element is 6, which corresponds to Carbon in the periodic table.
Therefore, the missing element is Carbon-14 (^14/6C).
c. 0/-1e + 76/36Kr -> ?
In this equation, an electron (^0/-1e) is added to Krypton-76 (^76/36Kr).
Since electrons have no mass or atomic number, adding an electron does not change the atomic number or mass number of an atom. Thus, the missing element should also be Krypton-76 (^76/36Kr).
d. 4/2He + 241/95Am -> ? + 2 1/0 n
Here we have a Helium-4 (^4/2He) nucleus colliding with an Americium-241 (^241/95Am) nucleus. The result is the formation of an unknown element and two neutrons (^2 1/0 n).
To balance the atomic numbers, Helium has an atomic number of 2 and Americium has an atomic number of 95. The sum of the atomic numbers on the right side should be equal to 97 (2 + 95).
Now let's balance the mass numbers. Helium has a mass number of 4 and Americium has a mass number of 241. The sum of the mass numbers on the right side should be equal to 245 (4 + 241).
Combining this information, the unknown element should have an atomic number of 97 and a mass number of 245.
Therefore, the missing element is Einsteinium-245 (^245/97Es).
Remember, balancing nuclear equations involves conserving both atomic numbers and mass numbers.