The energy required to separate all the nucleons inside a nucleus is called binding energy of the nucleus. Which nucleus posses maximum binding energy per nucleon?

In the given question we have to use the concept of Binding Energy per Nucleon versus Mass Number Graph and apply it to solve the given question. At the peak of binding energy, nickel-62 is the most tightly bound nucleus (per nucleon), followed by iron-58 and iron-56. Hence, nickel-62 will be the correct answer of the given question.

The energy required to separate all the nucleons inside a nucleus is called binding energy of the nucleus. Draw the graph showing the variation of binding energy per nucleon as a function of mass number.

In the given question we have to use the concept of and apply it to solve the given question. The variation of binding energy per nucleon versus mass number is shown in figure. Inferences from graph Hence, above graph will be the correct answer of the given question.

The energy required to separate all the nucleons inside a nucleus is called binding energy of the nucleus. Write an expression for binding energy in terms of mass defect.

In the given question we have to use the concept of binding energy and mass defect and the relation between them and apply it to solve the given question. Consider a nucleus Z A X that has Z protons and (A-Z) neutrons. ∆ m = Z m p - A - Z m n - m N w h e r e m n = m a s s o f n e u t r o n m p = m a s s o f p r o t o n m N = m a s s o f n e u c l e u s The expression of binding energy in terms of mass defect is ∆ E b = ∆ m × c 2 Hence, ∆ E b = ∆ m × c 2 will be the correct answer of the given question.

What is meant by half life of a radioactive substance? Write its relation with decay constant.

Taken for the number of radio active nuclei to disintegrate to half of its original number of nuclei is called Half life period. Relation between half the period and decay constant : The radioactive decay law N = N 0 e – λt States that the number of radioactive nuclei in a radioactive sample decreases exponentially with time. Here λ is called decay constant. If N 0 is the number of nuclei at t = 0 and N is the radioactive nuclei at any instant of time t . Substituting N = N ∘ 2 at t = T in N = N 0 e – λt . Where T is half life of the radioactive substance. N 0 2 = N 0 e – λt e λT = 2 ; λT = ln 2 T = ln 2 λ = 2 . 303 log 10 2 λ ∴ T = 0 . 693 λ The above equation represents the relation between half life ( T ) and decay constant ( λ ) . ' λ ' is different for different radioactive substances. Hence, it is the required answer.

Spontaneous and continuous disintegration of a nucleus of a heavy element with the emission of certain types of radiation is known as radioactivity. The radioactive isotope ' D ' decays according to the sequence D → β - D 1 → α D 2 If the mass number and atomic number of D 2 are 172 and 71 respectively, what are the: i Mass number, ii Atomic number of D .

Let the mass number of the element D be A , and its atomic number be Z . Hence, the element can be denoted as: D Z A . In an alpha α decay, an unstable nucleus changes to other element by shooting out a particle composed of two protons and two neutrons. Now, analysing the sequence from its end, we get: D 1 → α D 2 71 172 G i v e n i n t h e q u e s t i o n ⇒ D 1 + He 2 4 → α D 2 71 172 ⇒ D 1 69 168 + He 2 4 → α D 2 71 172 In beta minus β - decay, a neutron is converted into a proton, and the process creates an electron and an electron antineutrino. Now analysing the first part of the decay sequence to get the mass and atomic number of the element D , we get: D Z A → β - D 1 69 168 ⇒ D Z A + e - 1 0 + v ¯ 0 0 → β - D 1 69 168 Hence, A + 0 + 0 = 168 ⇒ A = 168 , and Z - 1 + 0 = 69 ⇒ Z = 70 Hence, the mass number of the element D is 168 , and its atomic number is 70 .

Prove the relation N t = N 0 e - λA

If N o is the number of nuclei presents at the initial time and N is the number of nuclei presents at the time t then, ∫ N o N d N N o = − ∫ 0 t λ d t = − λ ∫ 0 t d t log e N − log e N o = − λ t log e N N o = − λ t N N o = e − λ t N = N o e − λ t , hence the expression for the radioactive decay is derived.