Physics M.C.Q [1M]

2. $\alpha$-practicle

Explanation:

The binding energy is a energy that holds the nucleus together. 

Thus, more binding energy will give more stable nuclei. Here alpha particle has more binding energy so it will be more stable than deutron.

1. Q₁ = (M_x - M_y)c₂ and Q₂ = (M_x - M_y - 2m_e)c².

Sloution:

Key concept: Q value or energy of nuclear reaction: The energy absorbed or released during a nuclear reaction is known as Q-value of nuclear reaction.

Q-value = (Mass of reactants – mass of products)c² Joules = (Mass of reactants – mass of products) amu

If Q < 0, the nuclear reaction is known as endothermic. (The energy is absorbed in the reaction).

If Q > 0, the nuclear reaction is known as exothermic. (The energy is released in the reaction).

Let the nucleus be _ZX^A.

$\beta$ decay is respresented as: $_\text{Z}\text{X}^\text{A}\rightarrow_{\text{z}+1}\text{Y}^{\text{A}}+_{-1}\text{e}^0+\overline{\text{v}}+\text{Q}_2$

Q₁ = [m_n(_zX^A) - m_n (_z+1Y^A) - m_e]c²

= [m_n (_zX^A) + Zm_e - m_n (_z+1Y^A) - (Z + 1)m_e]c²

= [m(_zX^A) - m(_z-1Y^A)]c²

⇒ Q₁ = (M_x - M_y)c²

$\beta^+$ decay is represented as; $_\text{Z}\text{X}^\text{A}\rightarrow_{\text{z}-1}\text{Y}^{\text{A}}+_{+1}\text{e}^0+{\text{v}}+\text{Q}_2$

Q₂ = [m_n(_zX^A) - m_n (_z-1Y^A) - m_e]c²

= [m_n (_zX^A) + Zm_e - M_n (_z-1Y^A) - (Z - 1)m_e - 2m_e]c²

= [m(_zX^A)' - m(_z-1Y^A) - 2m_e] c²

⇒ Q2 = (M_x - M_y - 2m_e)c².

2. Packing fraction

Explanation:

Packing fraction is the mass defect per nucleon i.e. elementary particle in the nucleus. The difference between atomic weight and mass number i.e. mass of elementary particle in the nucleus is known as mass defect. Hence,

$\text{p}=\frac{\Delta\text{m}}{\text{A}}$

$\text{p}=\frac{\text{M - A}}{\text{A}}$

3. 3 (three)

Explanation:

Carbon has three isotopes $12​\text{C}\\ \ 6$, $13​\text{C}\\ \ 6$ and $14​\text{C}\\ \ 6$

2. Mass defect

Explanation:

The difference between the sum of the masses of the constituent particles and the mass of an atom is called mass defect.

The energy released due to mass defect was given by Einstein

E = $△$mc²

4. 7MeV

Explanation:

BE)_Hc​ = 28MeV, A of He = 4

BE)_He​ = $\frac{28}{4}$​MeV

= 7 MeV

2.  F_pp = F_pn = F_nn

Explanation:

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other. These forces are equal in magnitude, irrespective of the charge present on the nucleons.

$\therefore$ F_pp = F_pn = F_nn

Here, F_pp = F_pn = F_nn denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

1.  A straight line.

Explanation:

The average nuclear radius (R) and the mass number of the element (A) has the following relation:

$\text{R}=\text{R}_{0}\text{A}^{\frac{1}{2}}$

$\frac{\text{R}}{\text{R}_{0}}=\text{A}^{\frac{1}{3}}$

In $\Big(\frac{\text{R}}{\text{R}_0}\Big)=\frac{1}{3}$ In A

Therefore, the graph of ln$\Big(\frac{\text{R}}{\text{R}_0}\Big)$ versus ln A is a straight line passing through the origin with slope $\frac{1}{3}.$

2. C−12

Explanation:

Modern atomic weight scale is based on C−12.

The standard unit for expressing the mass of atom is amu (atomic mass unit).

It is equal to $\frac{1}{12}$ of the mass of an atom of carbon-12 equal to 1.6605×10⁻¹⁹g.

amu is also called as avogram.

Avogram is a unit of mass and weight equal to one gram divided by the Avogadro's number.

2. $\beta^-$decay

Explanation:

$\beta^-$ decay: $_\text{Z}^\text{A}\text{​X}→_{\text{Z}+1}^{\text{A}}\text{​Y} + _{−1}^0\text{​e}$

Thus, the atomic number is increased in $\beta^-$ decay.

1. Triton energy is less than that of a He³ nucleus.

Solution:

The nucleus of Tritium (₁H³) contains 1 proton and 2 neutrons.

A neutron decays as $\text{n}\rightarrow\ \text{p}+\overline{\text{e}}+\overline{\text{v}},$ the nucleus may have 2 protons and one neutron i.e., tritium will transform into ₂He³. It means triton energy is less that of ₂He³ nucleaus. Tirton energy is less than that of ₂He³ nucleus, which simple mean transformation is not allowed energetically.

3. More than half the active nuclei decay.

Explanation:

The average life is the mean life time for a nuclei to decay.

It is given as

$\tau=\frac{1}{\lambda}=\frac{\text{T}_{\frac{1}{2}}}{0.693}$

Here, $\tau,\lambda$ and $\text{T}_{\frac{1}{2}}$ are the average life, decay constant and half life-time of the active nuclei, respectively. The value of the average lifetime comes to be more than the average lifetime. Therefore, in one average life, more than half the active nuclei decay.

3. Decay constant of A is greater than that of B but it does not always decay faster than B.
4. Decay constant of B is smaller than that of A but still its decay rate becomes equal to that of A at a later instant.

Solution:

It can be observed from the figure that the slope of curve A is greater that thet of curve B, it means the rate of decay is faster for A than that of B.

According to Rutherford and Soddy law for radioactive decay $-\Big(\frac{\text{dN}}{\text{dt}}\Big)\propto\lambda$, where decay,

Hence at point P, rate of decay for both A and B is the same.

4. 1

Explanation:

Deuterium: $\frac{2}{1}$​H

Mass number: A = 2

Atomic number (or number of protons), Z = 1

Number of neutrons, Nn​ = A − Z = 2 − 1 = 1

3. A neutron in the nucleus decays emitting an electron.

Explanation:

Negative beta decay is given as

$\text{n}\rightarrow\text{p + e}^-+\bar{\text{v}}$

Neutron decays to produce proton, electron and anti-neutrino.

1. $\alpha-\text{decay}$

2. $\beta^+-\text{decay}$

Explanation:

In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number (atomic number) Z as well as neutron number N by 2.

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}-4}_{\text{Z}-2}\text{Y}+\text{ }^4_2\text{He}$

During $\beta^--\text{decay},$ a neutron is converted to a proton​, an electron and an antineutrino. Thus, there is an increase in the atomic number.

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}+1}\text{Y}+\text{e}^-+\bar{\text{v}}$

During $\beta^+-\text{decay},$ a proton in the nucleus is converted to a neutron​, a positron and a neutrino in order to maintain the stability of the nucleus. Thus, there is a decrease in the atomic number. ​

$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}-1}\text{Y}+\beta^++\text{v}$

When a nucleus is in higher excited state or has excess of energy, it comes to the lower state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. The element in the gamma decay doesn't change.

Therefore, alpha and beta plus decay suffer decrease in atomic number.

2. M = m_proton + m_electron - $\frac{\text{B}}{\text{C}^2}$ (B = 13.6eV).

Solution:

Given, $\text{F}=\frac{\text{GMm}}{\text{r}^2}$

M = Effective mass of hydrogen atom = mass of electron + mass of proton - $\frac{\text{B}^2}{\text{C}}$

Where B is BE of hydrogen atom = 13.6eV.

3. $\frac{1}{12}$^th mass of one carbon-12 atom

Explanation:

One a.m.u. is defined as mass of $\frac{1}{12}$^th the mass of one carbon-12 atom.

2. De-Broglie wavelength associated with electron in $\beta - \text{decay}$ is much greater than the size of nucleus.

1. A neutron in the nucleus decays emitting an electron.

Explanation:

$\beta^-$ emission is due to decay of neutron in the nucleus n → p + e⁻.

2. Carbon

Explanation:

²²Ne decays

$\alpha$particle = He²⁺

Mass No = 4

p = 2, n = 2

So, New mass no. = 22 − 8 = 14

Atomic No. = 10 − 4 = 6

So, the new element is _6​C¹⁴

2. $40\text{K}\\19$