31 Year NEET Previous Year Questions: Nuclei - 2 - NEET MCQ

9750 = KN₀ ............. (1)
975 = KN ............. (2)
Dividing (1) by (2)

   = 0.4606 = 0.461 per minute

The nuclear radius R = R_oA^1/3

Binding energy
= 117 × 8.5 + 117 × 8.5 – 236 × 7.6
= 234 × 8.5 – 236 × 7.6
= 1989 – 1793.6 = 200 MeV
Thus, in per fission of Uranium nearly 200
MeV energy is liberated

X must have zero charge and almost zero mass as electron  is emitted. Hence X must be antineutrino.

The penetrating power of radiation is directly proportional to the energy of its photon.
Energy of a photon 

∴ Penetrating power 

λ is minimum for γ-rays, so penetrating power is maximum of γ-rays.

Moderator used in nuclear r eactor are graphite and heavy water.

80 = 20 × n_A ⇒ n_A = 4
80 = 40 × n_B ⇒ n_B = 2

Suppose that, The number of ¹⁰B type atoms = x and the number of ¹¹B type atoms = y
Weight of ¹⁰B type atoms = 10x
Weight of ¹¹B type atoms = 11y
Total number of atoms = x + y

∴  Atomic weight

⇒10x + 11y = 10.81x + 10.81y
⇒ 0.81x = 0.19y

We know that alpha particles are the nucleus of ionized helium atoms which contain two protons and two neutrons. These are emitted by the nuclei of certain radioactive substances. Streams of alpha particles, called α-rays, produce intense ionisation in gases through which they pass and are easily absorbed by matter.

Emission of 1α particle led to decrease in atomic number by 2 while mass number by 4.
On the other hand, emission of 2β particles increases  atomic number by 2. Hence, overall emission of, 1α and 2β particles led to decrease in mass number by 4.

when  

Atomic hydrogen is unstable and it has life period of a fraction of a second.

According to liquid drop model of nucleus, an excited nucleus breaks into lighter nuclei just like an excited drop breaks into tiny drops.

X is a neutrino, when β-particle is emitted.

Given : Mass of neutron = M_n Mass of proton = M_p; Atomic mass of the element = M ;
Number of neutrons in the element = N and number of protons in the element = Z.
We know that the atomic mass (M) of any stable nucleus is always less than the sum of the masses of the constituent particles.
Therefore, M < [NM_n + ZM_p].

Atoms decayed  

 = 3.5 × 10¹⁶

For nuclear fusion process the nuclei with low mass are suitable.

As a result of fusion, enormous amount of heat is liberated which is the main cause of source of solar energy.

-1^e0 represents a β-decay..

Using the relation for mean life.
Given : 

Then from 

In case of hydrogen atom, Mass number = atomic number

Δm = (2 × 1.0074 + 2 × 1.0087 – 4.0015) = 0.0307
E = (Δm) × 931 MeV = 0.0307 × 931 = 28.5 MeV

Hence, V_atom = 10¹⁵ × V_nucleus

m₃ < (m₁ + m₂) (∵ m₁ + m₂ = m₃ + E ]
as E = [m₁ + m₂ – m₃] C²

has Z protons and (A – Z) neutrons

Mass defect 

Mass of nucleus = Mass of proton + mass of neutron  – mass defect

Isotones means equal number of neutrons i.e.,   (A–Z) = 74 –34 = 71 – 31 = 40.