The behavior of radioactive nuclei is different from that of normal nuclei. Let’s suppose you take two weighing pans; you put carbon in one, radioactive uranium in another in suitable conditions. After some time has passed, you would notice that the mass of the carbon remains unchanged, but the mass of uranium has reduced. Now why does this happen? This is due to a phenomenon called radioactive decay or radioactivity; a phenomenon which was intensively experimented upon by scientists like Madam Curie and Henry Becquerel. Due to radioactivity, uranium will slowly decay; when you bring a Geiger-Counter near it (a device used to detect radioactivity), the counter shows deflection, meaning that radiation is emitted from the Uranium. Radioactive decay is of many types. It mainly depends upon the parent nucleus undergoing decay. In this article, we will explore alpha decay in detail.
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ZXA ━━━⟶ Z-2YA-4 + 2He4
Mass of the parent nucleus, m X = m( ZXA) - Zme
Mass of the daughter nucleus, mY = m( Z-2YA-4) - (Z - 2)me
Mass of the alpha particle mα = m ( 2He4) - 2me, me - mass of the electron in amu.
The energy liberated during a nuclear reaction is called its Q-value. The Q- value of the - decay can be calculated as, c- velocity of light.
92U238 ━━━⟶ 90Th234 + 2He4
The total momentum is conserved in an alpha decay reaction. Let pi indicate initial momentum and pf indicate the final momentum. Since the parent nucleus is at rest, pi=0.
ZXA ━━━⟶ Z-2YA-4 + 2He4
Let m indicate the mass of the -particle and my- mass of the daughter nucleus. Let v indicate the recoil velocity of the -nucleus and vy indicate the velocity of the daughter nucleus. Then final momentum pf = mαvα-myvy.
The α- particle moves in the direction opposite to that of the daughter nucleus.
Q1) Complete the following reaction
88Ra226 ━━━⟶ 86Rn222 + ?
Solution) Since there is a decrease in atomic number by 2 and mass number by 4, the above reaction is an - decay.
⇒ 2He4 is emitted in the process.
∴ 88Ra226 ━━━⟶ 86Rn222 + 2He4
Q2) The total number of and particles emitted in the following nuclear reaction is:
92U238 ━━━⟶ 82Pb214 is
92U238 ━━━⟶ 82Pb214 + 6 2He4 + 2e-1
No. of alpha particles = 6
No. of beta particles = 2
Q3) A nucleus having a mass number of 220, initially at rest, emits an α - particle. If the Q - a value of the reaction is 5.5 MeV, then the energy of the α - particle will be
(a) 4.8 MeV (b)5.4 MeV ( c ) 5.5 MeV (d)6.8 MeV
Let X ━━━⟶ Y+ 2He4
Given, mass of X = 220
∴ Mass of Y = 220 - 4 = 216
Let K1 be the kinetic energy of Y and K2 be the kinetic energy of the 2He4 nucleus.
Then K1 + K2 = 5.5 MeV -- (i)
Now linear momentum , hence momentum p1 of Y can be written as
And linear momentum of particle can be written as p2 =
Since momentum is conserved ; p1=p2
= =54 ; K2=54 = K1 -- (ii)
Solving (i) and (ii), we get K2 = 5.4 MeV.
Q4) Alpha particles emitted from a radioactive substance are
(a)Ionized hydrogen nuclei
(b)Doubly ionized helium atom
(c)Uncharged particles having mass equal to proton
Alpha particles are doubly ionized helium atoms.
Question 1. Heavy nuclei like uranium do not undergo β - decay. Why?
Answer. Heavy nuclei like Uranium undergo α - decay. β - decay occurs in lighter nuclei which have excess protons or neutrons
Question 2. Write the - decay reaction of 88Ra226?
Answer. 88Ra226 ━━━⟶ 86Rn222 + 2He4
Question 3. Write the - decay reaction of 92U238 to form thorium?
Answer. 92U238 ━━━⟶ 90Th234 + 2He4
Question 4. Write the unit of Q- value?
Answer. The unit of Q- value is MeV.