In the hypothetical fission reaction X 92 236 → Y a 144 + Z 36 b + 3 n 0 1 What are the values of a and b ?

Given reaction, X 92 236 → Y a 144 + Z 36 b + 3 n 0 1 From the atomic mass conservation 236 = 144 + b + 3 ⇒ b = 89 From the charge conservation 92 = a + 36 + 0 ⇒ a = 56

In the hypothetical fission reaction X 92 236 → Y a 144 + Z 36 b + 3 n 0 1 What are the values of a and b ?

Given reaction, X 92 236 → Y a 144 + Z 36 b + 3 n 0 1 From the atomic mass conservation 236 = 144 + b + 3 ⇒ b = 89 From the charge conservation 92 = a + 36 + 0 ⇒ a = 56

What is half-life of a radioactive nucleus? Show that half-life is inversely proportional to the decay constant of the radioactive sample.

In radioactive nucleus, half-life is the time taken by the nuclei to decay half of it. Here, the rate of decay remains constant throughout the decay process. So, rate of decay: d N d t = - λ N ⇒ d N N = - λ d t . Here, λ is decay constant. By integrating both sides: ∫ N ∘ N 1 N d N = - λ ∫ 0 t d t ⇒ ln N N ∘ = - λ t ⇒ N = N ∘ e - λ t At half life time: t = τ 1 2 the value of N = N ∘ 2 . By substituting both in the above equation: N ∘ 2 = N ∘ e - λ τ 1 2 ⇒ 2 = e λ τ 1 2 ⇒ ln 2 = λ τ 1 2 ⇒ 0 . 633 τ 1 2 = λ Hence, half-life is inversely proportional to the decay constant of the radioactive sample.

Find the binding energy per nucleon of an a-particle in MeV . (Take, 1 a . m . u . = 931 . 5 MeV ) Given- mass of α -particle = 4 . 00150 a . m . u . mass of proton = 1 . 00728 a . m . u . mass of neutron = 1 . 00867 a . m . u .

The binding energy: E B = ∆ m c 2 mass of α -particle m α = 4 . 00150 a . m . u . mass of proton m p = 1 . 00728 a . m . u . mass of neutron m n = 1 . 00867 a . m . u . Here, m α ∆ m = m α - 2 m p + m n = - 0 . 0304 a . m . u . By, substituting the values: E B = ∆ m c 2 = 0 . 0304 a . m . u . × 931 . 5 MeV = 28 . 3176 MeV Binding energy per nucleon will be, E B p + n = 28 . 3176 MeV 4 = 7 . 08 MeV Hence, the total energy per nuclide is 7 . 08 MeV .

A → α A 1 → β A 2 → α A 3 The mass number and atomic number of A are 180 and 72 respectively. What are these numbers for A 3 ?

Atomic number of A is 72 and its atomic mass number is 180 . After α -decay of A , atomic number and atomic mass number of A 1 is 70 and 176 respectively. After β -decay of A 1 , atomic number and atomic mass number of A 2 is 71 and 176 respectively. After α -decay of A 2 , atomic number and atomic mass number of A 3 is 69 and 172 respectively. Hence, the atomic number and atomic mass number of A 3 is 69 and 172 respectively.

Two or more nuclides having the same neutron number but different number of protons are said to be isotones. For example S 36 , 37 U , 38 A r , 39 K , 40 C a all contain 20 neutrons.

State radioactive decay law. Derive the relation N = N 0 e - λ t , where symbols have their usual meanings.

According to radioactive decay law, the rate of disintegration of a radioactive substance at an instant is directly proportional to the number of nuclei in the Radioactive substance at that time. R = - dN dt or - dN dt ∝ N ⇒ R = - dN dt = λN Where λ is the constant, called the decay constant. The equation can be written as d N N = - λ d t Integrating, we have ∫ d N N = - λ ∫ d t l o g e N = - λ t + C … ( i ) Where C is the constant of integration, at t = 0 ⇒ N = N 0 Therefore, C = log e N 0 Now, form equation (i) log e N = - λt + log e N 0 log e N - log e N 0 = - λt log e N N 0 = - λt N N 0 = e - λt ⇒ N = N 0 e - λt