alpha decay and nuclear reactions help
1.
___ have unstable nuclei and emit rays and particles to become more stable. this process was discovered by ____ which she called _____.
fill in the blanks.
2.
write the symbol produced when radon-222 undergoes alpha decay
3.element produced when strontium-90 undergoes beta decay?
4.complete the reaction
238 / 92 U -> 234 / 90 TH + ?
5.
26 / 12 Mg -> 26 / 13 AI + ?
6.fill in blanks
210 / ? Pb -> ? / 83 Bi + o / -1e
7. half life. write the # of grams remaining after each half life 120 g is the starting amount
amount after
3 hours
6 hours
9 hours
for number one i got radioactive atoms, and then marie curie and the third blank radioactivity.for the rest of the questions there’s blank spots basically a fraction and then a spot for a letter. for number two i got 218 / 84 Po. for three i got 90 / empty Y. for four i got 4 / 2 a. the rest i have not gotten
For #1 I agree that Curie SOUNDS right, especially when the rest of sentence says "she"; however, my understanding is thatHenri Becquerel was given credit for discovering radioactivity. For #3, Sr90 goes to Y90. I worked 2 and 4 for you.
okay thankyou i have answers for all of them now except for the final one. it’s a radioisotope has a half life of three hours how much of a 120g sample remains after 9 hours. and i need to figure out for 3 hours 6 hours and 9 hours
half life of 3 hours means half is gone after three hours; therefore, after 3 hrs you will hve 120/2 = 60. After 6 hrs another half will be gone and after 9 another half will be gone
60/2 = 30
30/2 = ?
This is easy because the problem has times that are even numbers of a half life. If they asked you for how much is left after 3.5 hours it makes it a bit harder. Here is how you do those. First you determine the constant. That is k = 0.693/half life or k = 0.693/3 =0.231, then
ln(No/N) = kt where N is how much is left. No is how much you started with. k is from above. t is time in hrs since the half life is in hrs. So we do 3 hrs.
ln(120/N) = 0.231*3 = 0.693
120/N = 2
120 = 2N and N = 120/2 = 60
Plug in different times of 6 and 9 hrs etc and any other odd number hrs like 3.5 hrs or 11.9 hrs. Good luck.
1. Atoms have unstable nuclei and emit rays and particles to become more stable. This process was discovered by Marie Curie, which she called radioactivity.
2. The symbol produced when radon-222 undergoes alpha decay is 218/86 Rn.
3. The element produced when strontium-90 undergoes beta decay is yttrium-90 (90/39 Y).
4. Completing the reaction:
238/92 U -> 234/90 Th + 4/2 He (alpha particle)
5. 26/12 Mg -> 26/13 Al + 0/-1 e (beta particle)
6. Filling in the blanks:
210/82 Pb -> 210/83 Bi + 0/-1 e (beta particle)
7. Half-life and grams remaining:
After 3 hours: Approx. 60 g remaining
After 6 hours: Approx. 30 g remaining
After 9 hours: Approx. 15 g remaining
Remember, these answers are based on simplified examples. Always consult reliable sources for accurate information regarding nuclear reactions.
1. Elements have unstable nuclei and emit rays and particles to become more stable. This process was discovered by Marie Curie, which she called radioactivity.
2. To find the symbol produced when radon-222 undergoes alpha decay, we need to understand the process of alpha decay. Alpha decay occurs when an alpha particle, which consists of two protons and two neutrons, is emitted from the nucleus of the parent element. In the case of radon-222, the alpha decay would be represented as follows:
Radon-222 (symbol: Rn-222) -> Alpha particle (symbol: He-4) + Daughter element
So, after alpha decay, the symbol produced would be He-4.
3. To find the element produced when strontium-90 undergoes beta decay, we need to understand the process of beta decay. Beta decay occurs when a neutron in the nucleus is converted into a proton, and an electron (beta particle) and an antineutrino are emitted. In the case of strontium-90, the beta decay would be represented as follows:
Strontium-90 (symbol: Sr-90) -> Beta particle (symbol: e- or β-) + Daughter element
So, the element produced would depend on the specific daughter element resulting from the beta decay.
4. To complete the reaction:
238 / 92 U -> 234 / 90 Th + ?
We can see that the total mass number on both sides of the equation is conserved (238 = 234 + mass of unknown particle). Similarly, the total atomic number is also conserved (92 = 90 + atomic number of unknown particle).
Based on this conservation, we can deduce that the unknown particle has a mass number of 4 (238 - 234 = 4) and an atomic number of 2 (92 - 90 = 2). This matches the characteristics of an alpha particle, which has a mass number of 4 and an atomic number of 2.
So, the completion of the reaction would be:
238 / 92 U -> 234 / 90 Th + 4 / 2 He
5. To determine the product when 26 / 12 Mg undergoes alpha decay:
26 / 12 Mg -> 26 / 13 Al + ?
In alpha decay, an alpha particle with a mass number of 4 and an atomic number of 2 is emitted from the nucleus. So, the completion of the reaction would be:
26 / 12 Mg -> 26 / 13 Al + 4 / 2 He
6. To determine the missing numbers:
210 / ? Pb -> ? / 83 Bi + 0 / -1e
In this beta decay reaction, a beta particle (electron) is emitted. The beta particle has an atomic number of -1 and a mass number of 0. In this case, the mass number of the parent nucleus (210) decreases by 1, while the atomic number increases by 1.
Using this information, we can determine the missing values:
210 / 82 Pb -> 210 / 83 Bi + 0 / -1e
7. Half-life refers to the time it takes for half of the initial amount of a radioactive substance to decay. To find the amount remaining after each half-life, we need to know the half-life of the substance.
Let's assume the half-life of the substance is 1 hour:
- After 3 hours: 120 g -> 60 g (half of 120 g, one half-life)
- After 6 hours: 60 g -> 30 g (half of 60 g, two half-lives)
- After 9 hours: 30 g -> 15 g (half of 30 g, three half-lives)
Note that the half-life value will vary depending on the specific radioactive substance being considered.
I'm not about to spend the time to do all of the homework for you; however, I will get you started. If you have specific questions about my answers or one of your questions you should post a follow up.
I can't help with #1. Do you have choices for the blank spaces.
To do #2 and most of the others, remember to make the atomic numbers add up on both sides and make the mass number add up on both sides. Here is how you do it for #2.
The upper number is the mass number and is the sum of the protons and neutrons.
The lower number is the atomic number and is the + charge, due to number of protons in the nucleus. Radon is atomic number 86 and mass number 222. So if it ejects an alpha particle, which is a helium atom or 4/2He
.........222/86Rn ==> 4/2He + X. You need to determine what X is and the mass number and atomic number of X. Again, make the upper numbers and lower numbers end up balanced on both sides of th equation.
Note that on the left we have 222. On the right we have 4 + ? = 222 and ? must be 218. On the left we have 86 and on the right we have 2 + ? = 86 so ? must be 84. so we have
..........222/86Rn ==> 4/2He + X. which from the above becomes
..........222/86Rn ==> 4/2He + 218/84X. Now we must determine the element X. It has an atomic number of 84 with a mass number 218. A look at the periodic table tells you element number 84 is Po. The complete equation then becomes
..............222/86Rn ==> 4/2He + 218/84Po
For #4 238 / 92 U -> 234 / 90 Th + ? Just remember, make upper and lower numbers add up. 238/92U -> 234/90Th + 4/2X. So X is the particle with atomic number 2 and mass number of 4 which is an alpha particle.
For #6. 210/? Pb -> ?/83Bi + o/-1e
Look on the periodic table. Pb is 82.
210/82Pb -> ?/83Bi + o/-1e
?/83Bi. That must be 210 = ? + o/-1e so ? must be 210
So 210/82Pb ==> 210/83Bi + o/-1e
Remember that o/-1e us a beta minus particle.
This works three for you and shows how it's done. I'll leave the others for you. Post here if you would like for me to check your answers.