31 Year NEET Previous Year Questions: Nuclei - NEET MCQ

requires a and b amount of energies for their nucleons to be separated.
releases c amount of energy in its formation i.e., in assembling the nucleons as nucleus.
Hence, Energy released =c – (a + b) = c – a – b

Binding energy per nucleon for fission products is higher relative to Binding energy per nucleon for parent nucleus, i.e., more masses are lost and are obtained as kinetic energy of fission products. So, the given ratio < 1.

Binding energy per nucleons decreases with increase in mass numbers.
The decrease of binding energy per nucleon for nuclei with a high mass number is due to increased Coulombic repulsion between protons inside the nucleus.

Energy released = 28 – 2 × 2.2 = 23.6 MeV (Binding energy is energy released on formation of Nucleus)

In the case of formation of a nucleus, the evolution of energy equal to the binding energy of the nucleus takes place due to the disappearance of a fraction of the total mass. If the quantity of mass disappearing is ΔM, then the binding energy is

BE = ΔMc²
From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write.

ΔM= [ZM_p + (A-Z)M_n -M(A,Z)
 

Where M (A, Z)  is the mass of the atom of mass number A and atomic number Z. Hence, the binding energy of the nucleus is

BE = [ZM_p + (A-Z)M_n -M(A,Z)]c²

Where N = A -Z = number of neutrons.

It has been known that a nucleus of mass number A has radius R = R₀A^1/3,
where R₀ = 1.2 × 10^–15m and A = mass number

In case of  let nuclear radius be R₁

and for  nuclear radius be R₂

For

For

Mass defect = ZM_p + (A –Z)M_n–M(A,Z)

  ZM_p + (A–Z) M_n–M(A,Z)

∴ M (A, Z) = ZM_p + (A–Z)M_n–

Requird ratio of nuclear densities =

Binding energy of two ₁H² nuclei = 2(1.1 × 2) = 4.4 MeV
Binding energy of one ₂He⁴ nucelus = 4 × 7.0 =28 MeV
∴ Energy released = 28 – 4.4 = 23.6 MeV

E = mc²

So, mass decay per second

(Power in watt)

and mass decay per hour =

= 4 × 10^–8 kg

= 40 microgram

Momentum

Recoil energy

When the coulomb repulsion between the nuclei is overcome then nuclear fusion reaction takes place. This is possible when temperature is too high.

Initially  P → 4N₀  
             Q → N₀
Half life T_P = 1 min.
T_Q = 2 min.
Let after time t number of nuclei of P and Q are equal, that is

⇒ t = 4 min

at t = 4 min.

or population of R

The radius of the nuclears is directly proportional to cube root of  atomic number i.e. R ∝ A^1/3 ⇒ R = R₀ A^1/3, where R₀ is a constant of proportionality

where R₁ = the radius of ²⁷Al, and A₁ = Atomic  mass number of Al R₂ = the radius of ⁶⁴Cu and A₂ = Atomic mass number of C₄

Let, the amount of the two in the mixture will become equal after t years.
The amount of A₁, which remains after t years

The amount of A₂, which remains, after t years

According to the problem N₁ = N₂

t = 40 s

        ...(i)

     ...(ii)

Dividing equation (i) by equation (ii)

λ (t₂ - t₁) = ln2

 = 50 days

Mass defect Δm = 0.02866 a.m.u.
Energy = 0.02866 × 931 = 26.7 MeV

Energy liberated per a.m.u = 13.35/2 MeV        
= 6.675 MeV

From the graph of BE/A versus mass number A  it is clear that, BE/A first increases and then decreases with increase in mass number.