3. Radioactive decay of nuclei often involves several decays before a stable nucleus is formed. This is called a decay chain. What stable isotope is formed when radon-222 undergoes a decay chain of four alpha decays followed by four beta decays?
(A) tungsten-206
(B) platinum-206
(C) lead-206
(D) tungsten-214
(E) lead-214
4. Which nucleus balances the following nuclear equation for the fission of uranium-23? 235/92 U + 1/0 n -> 90/38 Sr + A/Z X + 2 1/0 n + y
(A) 146/54 Xe
(B) 146/52 Te
(C) 144/52 Te
(D) 144/54 Xe
(E) 142/50 Sn
Any help would be greatly appreciated; I don't understand this AT ALL. Please and thank you!
four alpha means 4*2+) have left
Rn222 is 86X222
so take away 4*mass of alpha, and 4 charge of alpha, you get
78X214, and element 78 is Pt, and weight
then you have 78Pt204, then four beta decays leave the mass as 204, but changes the atomic number to 82
82Pt204
To determine the stable isotope formed when radon-222 undergoes a decay chain of four alpha decays followed by four beta decays, we need to trace the decay chain and identify each decay step.
First, radon-222 decays through an alpha decay, which means it loses an alpha particle (2 protons and 2 neutrons). This results in the formation of a new nucleus.
Next, the newly formed nucleus also undergoes an alpha decay, and the process repeats for a total of four alpha decays. Each alpha decay leads to the formation of a different nucleus.
After the four alpha decays, the resulting nucleus undergoes four beta decays. In a beta decay, a neutron converts into a proton, emitting an electron (beta particle) and an antineutrino.
To determine the final stable nucleus formed, we need to identify the product of each decay step and continue the decay chain.
Starting with radon-222, we have the following steps:
Radon-222 -> Alpha Decay -> Polonium-218
Polonium-218 -> Alpha Decay -> Lead-214
Lead-214 -> Alpha Decay -> Bismuth-210
Bismuth-210 -> Alpha Decay -> Thallium-206
Thallium-206 -> Beta Decay -> Lead-206
Therefore, the stable isotope formed after the decay chain is Lead-206. Option (C) is the correct answer.
Now let's move on to solving the second question.
In the nuclear equation for the fission of uranium-235 (235/92 U), we have:
235/92 U + 1/0 n -> 90/38 Sr + A/Z X + 2 1/0 n + y
To balance the equation, we need to determine the values of A and Z for the nucleus A/Z X.
From the equation, we can see that 2 neutrons (2 1/0 n) are produced. This means the sum of neutrons in the products must be equal to the sum of neutrons in the reactants.
235/92 U: Neutrons = 235 - 92 = 143
2 1/0 n: Neutrons = 2
Sum of neutrons in the products = 143 + 2 = 145.
Now let's consider the atomic number (Z):
90/38 Sr: Z = 38
A/Z X: Z = Z
2 1/0 n: No protons
Sum of protons in the products = Z + 0 = Z.
Since the sum of protons in the reactants is 92 (235/92 U), the sum of protons in the products must also be 92.
Sum of protons in the products = 92.
From the atomic number, it can be deduced that Z = 92.
Now, let's determine the mass number (A) for A/Z X:
235 (mass number of U) = A + 2 (neutrons)
235 = A + 2
A = 233
Therefore, the nucleus A/Z X is 233/92 X. Option (A) 146/54 Xe is the correct answer.
To find the stable isotope formed when radon-222 undergoes a decay chain of four alpha decays followed by four beta decays, we can follow the decay steps and determine the resulting isotope at each step.
Starting with radon-222 (Radon-222 → Decay 1), it undergoes an alpha decay to produce polonium-218.
Polonium-218 (Decay 2) undergoes another alpha decay to produce lead-214.
Lead-214 (Decay 3) undergoes yet another alpha decay to produce bismuth-210.
Bismuth-210 (Decay 4) goes through alpha decay again to produce thallium-206.
Finally, thallium-206 (Decay 5) undergoes four beta decays to produce lead-206, which is a stable isotope.
Therefore, the stable isotope formed after these decay steps is lead-206.
So, the answer to question 3 is (C) lead-206.
Moving on to question 4, we have the fission of uranium-235. The equation provided is:
235/92 U + 1/0 n -> 90/38 Sr + A/Z X + 2 1/0 n + y
In this equation, 235/92 U represents the uranium-235 nucleus, 1/0 n represents a neutron, 90/38 Sr represents strontium-90, A/Z X represents the unknown product nucleus, 2 1/0 n represents two neutrons, and y represents some form of radiation (such as gamma radiation).
In this fission reaction, the sum of the mass numbers and atomic numbers on both sides of the equation should be equal to each other to maintain a balanced nuclear equation.
The mass number on the left side is 235 (from uranium-235) and the mass number on the right side is 90 (from strontium-90) + A (from the unknown product nucleus) + 2 (from two neutrons). Therefore, the sum of the mass numbers on the right side is 90 + A + 2.
Similarly, the atomic number on the left side is 92 (from uranium-235) and the atomic number on the right side is 38 (from strontium-90) + Z (from the unknown product nucleus). Therefore, the sum of the atomic numbers on the right side is 38 + Z.
Based on the above information, we can conclude that A + 2 = 235 and Z = 92 - 38.
Solving A + 2 = 235 gives A = 233.
Substituting the value of A into Z = 92 - 38 gives Z = 54.
Therefore, the unknown product nucleus represented by A/Z X is 233/54 Xe.
So, the answer to question 4 is (A) 146/54 Xe.
I hope this helps you understand the questions! Let me know if you have any further queries.