31 Year NEET Previous Year Questions: Nuclei - 3 - NEET MCQ

In two half lives, the activity becomes one fourth.
Activity on 1–8–91 was 2 micro–curie
∴ Activity before two months, 4 × 2 micro-curie = 8 micro curie

Two successive β decay increase the charge number by 2. Therefore, (d) is correct.

No. of half lives, 

From eqn. (ii), B has 2 units of charge more than C.
From eqn. (i), A loses 2 units of charge by emission of alpha particle. Hence, A and C are isotopes as their charge numbers are same.

K.E. of fastest electron = E – W₀ = 6.2 – 4.2 = 2.0 eV
= 2 × 1.6 × 10^–19 = 3.2 × 10^{– 19} J

Curie is a unit of radioactivity.

Average B.E./nucleon in nuclei is of the order of 8 MeV.

As ₆C¹³ and ₇N¹⁴ have same no. of neutrons (13 – 6 =7 for C and 14 – 7 =7 for N), so they are isotones.

Nuclear forces are short range attractive forces which balance the repulsive forces between the protons inside the nucleus.

Nucleus contains only neutrons and protons.

Z = 11 i.e., number of protons = 11, A = 23
∴ Number of neutrons = A – Z = 12
Number of electron = 0 (No electron in nucleus) Therefore 11, 12, 0.

Nuclear force is the same between any two nucleons.

Fusion reaction.

1 a.m.u = 931 MeV

α-particle = ₂He⁴ . It contains 2p and 2n.
As some mass is converted into B.E., therefore, mass of α particle is slightly less than the sum of the masses of 2p and 2n.

The nuclear radius r varies with mass number A according to the relation
r ∝ A^1/3 or A ∝ r³

Now,  density 

Further, mass ∝ A and volume ∝ r³

 constant

Increase of charge number. by 1 indicates β emission. Decrease of mass number by 4 and charge number by 2 indicates α emission.
No change of mass number and charge number indicates γ-emission.

From B.E. curve, the curve reaches peak for
₂₆ Fe⁵⁶

Fission rate

= 1.56 × 1011 s^–1

Applying principle of energy conservation, Energy of proton = total B.E. of 2α– energy of Li⁷
= 8 × 7.06 – 7 × 5.6
= 56.48 – 39.2 = 17.28 MeV

Moderator slows down the neutrons to thermal energies.

Ion is an atom or a group of atoms that has lost or gained electrons having negative or positive charge respectively. Thus, the ions knocked out from hot surfaces is electrons.

We know that

Therefore, in this process,

200 = 4n + 168 or  n 

Also, 90 = 2n – m + 80
or,  m = 2n + 80 – 90 = (2 × 8 + 80 – 90) = 6.
Thus,  respective number of α and β-particles will be 8 and 6.

Half-life = 30 minutes; Rate of decrease (N) = 5 per second and total time = 2 hours = 120 minutes. Relation for initial and final count rate

Therefore, N₀ = 16 × N = 16 × 5 = 80 s^–1.

Applying law of conservation of momentum,
m₁v₁ = m₂v₂

Hence,