Atoms and Nuclei

$\mathrm{A}-\mathrm{f}$

Strong nuclear forces are those forces which acts between inside of the nucleus of an atom. It is strongest fundamental force in nature.

We have taken its strength as 1 according to question B-h

Weak nuclear forces are those forces which comes in play in conversion of neutron to proton or vice versa, through-$p$-deccay. It is $\frac{1}{10^{13}}$ in strength in comparison with strong nuclear force.

C-e

Electromagnetic forces act between the charge particles. It's strength with strong nuclear force is $1 / 100$ times.

D-iGravitational forces acts between bodies with masses. It is always attractive in nature. Its relative-strength to strong nuclear force is $\frac{1}{10^{39}}$ times.

Hence, (A-f, B-h, C-e, D-i) is correct order.

Given,

Energy of single uranium $=200 \mathrm{MeV}$

Power $=5 \mathrm{~mW}$

Using the formula

$ \begin{aligned} & \frac{E}{t}=P \\ \Rightarrow & P=\frac{n E}{t} \end{aligned} $

[where $n$ is number of fission reaction]

$ \begin{aligned} \frac{n}{t} & =\frac{P}{E} \\ & =\frac{5 \times 10^{-3}}{200 \times 10^6 \times 1.6 \times 10^{-19}} \\ & =1.56 \times 10^8 \text { fission per second } \end{aligned} $

Hence, number of fission reaction per second is $1.56 \times 10^8$.

The Lyman series's largest wavelength occurs when transition takeplace from $n_i=2$ to $n_i=1$ wavelength is given by

$ \begin{aligned} \frac{1}{\lambda} & =R\left[\frac{1}{n_j^2}-\frac{1}{n_i^2}\right] \\ \frac{1}{\lambda_{\max }} & =R\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3 R}{4} \end{aligned} $

for Lyman series $=\frac{4}{3} R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)$

For maximum wavelength of Balmer's series transition occur from $n_t=3$ to $n_f=2$

$ \left[\frac{1}{\lambda_{\max }}\right]_{\mathrm{Balmer}}=R\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=R\left[\frac{5}{36}\right] $

$ \left[\lambda_{\max }\right]_{\text {Balmer }}=\frac{36}{5} R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2) $

From Eqs. (i) and (ii)

$ \frac{\left[\lambda_{\max }\right]_{\text {Lyman }}}{\left[\lambda_{\max }\right]_{\text {Balmer }}}=\frac{4}{3} \times \frac{5}{36}=\frac{5}{27} $

Hence, ratio of $\lambda_{\max }$ for Lyman and Balmer is $\frac{5}{27}$.

Given: Half-life of substance $A$ is twice the half-life of substance $B$.

Let the half-life of $B$ be $T$. Then the half-life of $A$ will be $2T$.

We know that the number of undecayed nuclei after $n$ half-lives is given by

$ N = N_0 \left(\frac{1}{2}\right)^n $

We are told that we observe the substances after three half-lives of $A$.

Three half-lives of $A$ means time $= 3 \times (2T) = 6T$.

In this same time, substance $B$ will complete $6$ half-lives (because its half-life is $T$).

So, for substance $A$, number of nuclei remaining after $3$ half-lives:

$ N_A(\text{remaining}) = N_A \left(\frac{1}{2}\right)^3 $

For substance $B$, number of nuclei remaining after $6$ half-lives:

$ N_B(\text{remaining}) = N_B \left(\frac{1}{2}\right)^6 $

Given that after this time, the number of nuclei of $A$ and $B$ are equal, so

$ N_A\left(\frac{1}{2}\right)^3 = N_B\left(\frac{1}{2}\right)^6 $

Dividing both sides by $\left(\frac{1}{2}\right)^6$:

$ N_A \left(\frac{1}{2}\right)^3 \div \left(\frac{1}{2}\right)^6 = N_B $

$ N_A \left(\frac{1}{2}\right)^{3-6} = N_B $

$ N_A \left(\frac{1}{2}\right)^{-3} = N_B $

$ N_A \times 2^3 = N_B $

$ 8N_A = N_B $

So,

$ \frac{N_A}{N_B} = \frac{1}{8} $

Hence, $\frac{N_A}{N_B}$ is $\frac{1}{8}$.

Strong Nuclear force is the force acting inside nucleus (between nucleons) of an atom.

It is strongest force in nature fundamentally. Weak nuclear force is the force responsible for changing of neutron to proton or vice versa via beta-decay. It is not strong as compared to strong nuclear force.

The ratio of strong and weak nuclear force's relative strength is $10^{13}$.

Energy of $n$th level of H -atom is

$ E_n=\frac{-13.6}{n^2} \mathrm{eV}, \text { for } n=1, E_1=-13.6 \mathrm{eV} $

The general formula for an Hydrogen like atom's energy of $n$th level is

$ \begin{aligned} & E_n=\frac{-2 \pi m e^4 z^2}{n^2 h^2} \\ \Rightarrow \quad & E_n \propto m \end{aligned} $

Now, according to question if mass of electron is doubled, (For $n=1$ )

$ \begin{aligned} E_n & =2\left(\frac{-2 \pi m e^4 z^2}{n^2 h^2}\right) \\ E_1 & =2 \times-13.6 \mathrm{eV} \\ & =-27.2 \mathrm{eV} \end{aligned} $

Given,

Half life of $X=$ mean life of $Y$

Initially both have same number of atoms

Half life is given by, (for $X$ )

$ t_{1 / 2}=\frac{\ln 2}{\lambda_x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Mean life for $Y$

$ t=\frac{1}{\lambda_y}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

as $t_{1 / 2}$ of $X=t$ of $Y$

$ \begin{aligned} & \frac{\ln 2}{\lambda_x} =\frac{1}{\lambda_y} \\ \Rightarrow & (0.6393) \lambda_y =\lambda_x \\ \therefore \lambda_y & >\lambda_x \end{aligned} $

Higher is the decay constant, higher is the rate of decay.

Hence $y$ will decay faster than $X$.

Heavy water chemically it is written as $\mathrm{D}_2 \mathrm{O}$ (also called as deuterium oxide) act as moderator in a nuclear reaction because it slows down the neutrons produced during fission process.

Heavy water is an excellent moderator because it can work at high temperature and has low absorption cross-section for neutrons.

Thus, option (d) is the correct option.

We know that, radius of a nucleus of mass number $A$ is given as

$ \begin{aligned} & R \propto A^{1 / 3} \\ \Rightarrow \quad & \frac{R_2}{R_1}=\left(\frac{A_2}{A_1}\right)^{1 / 3} \end{aligned} $

$ \frac{2 R}{R}=\left(\frac{A_2}{27}\right)^{1 / 3} \Rightarrow 2=\left(\frac{A_2}{27}\right)^{1 / 3} $

Cubing both side, we get

$ 8=\frac{A_2}{27} \Rightarrow A_2=216 $

Thus, we see that nuclei having mass number 216 is heavier nuclei. Binding energy per nucleon is smaller for such heavier nuclei. Thus, it is likely to undergo fission reaction.

The given nuclear reaction is shown as

$ \begin{aligned} & { }_Z^A X \xrightarrow{\alpha-\text { decay }}{ }_{Z-2}^{a-4} P \xrightarrow{\beta^{-}-\text {decay }} \\ & { }_{z-1}^{A-4} Q \xrightarrow{\alpha-\text { decay }} { }_{z-3}^{A-8} R\end{aligned} $

No. of neutrons in the nucleus $R$

$ \begin{aligned} & =(A-8)-(Z-3) \\ & =A-8-Z+3 \\ & =A-Z-5 \end{aligned} $

In hydrogen spectrum for shortest wavelength of Balmer series, $n_1=2$ and $n_2=\infty$

$ \begin{array}{ll} \therefore & \frac{1}{\lambda_{B S}}=R\left(\frac{1}{2^2}-\frac{1}{\infty}\right) \\ \Rightarrow & \frac{1}{\lambda_{B S}}=\frac{R}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

For longest wavelength of Balmer series.

$ \begin{aligned} & & n_1 & =2 \text { and } n_2=3 \\ \therefore & & \frac{1}{\lambda_{B L}} & =R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ & \Rightarrow & \frac{1}{\lambda_{B L}} & =R\left(\frac{5}{36}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Subtracting Eq. (ii) from Eq. (i), we get

$ \begin{aligned} & \frac{1}{\lambda_{B S}}-\frac{1}{\lambda_{B L}}=\frac{R}{4}-\frac{5 R}{36} \\ \Rightarrow & \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{R}{4}-\frac{5 R}{36} \end{aligned} $

$ \Rightarrow \quad \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{R}{9} \Rightarrow R=\frac{9}{\lambda_1}-\frac{9}{\lambda_2} $

According to given situation,

$ X \rightarrow Y+\alpha $

∴ Mass defect, $\Delta m=m_x-m_y-m_\alpha$

∴ Net kinetic energy gained in the process $=$ energy released in nuclear reaction

$ \begin{aligned} & =\Delta m \times c^2 \\ & =\left(m_x-m_y-m_\alpha\right) c^2 \end{aligned} $

$ \begin{aligned} &\text { } E_1=188 \mathrm{MeV}=188 \times 1.6 \times 10^{-13}\\ &\text { Energy released }=N \times 188 \times 1.6 \times 10^{-13}\\ &\begin{aligned} & =6.02 \times 10^{23} \times 188 \times 1.6 \times 10^{-13} \\ & =1810.8 \times 10^{10} \\ & =1810 \times 10^{12} \mathrm{~J} \end{aligned} \end{aligned} $

When comparing the relative strengths of the gravitational force ($F_1$) and the weak nuclear force ($F_2$), the ratio $F_2 / F_1$ is approximately $10^{26}$.

Explanation

Gravitational Force ($F_1$): This is the weakest of the four fundamental forces in nature. It acts between bodies with mass and is responsible for the attraction between them.

Weak Nuclear Force ($F_2$): This force plays a crucial role in nuclear processes such as beta decay, where protons can transform into neutrons and vice versa. Despite its name, it is significantly stronger than gravity.

The comparative strength of the weak nuclear force to the gravitational force is around $10^{26}$. This reflects just how much weaker gravity is compared to the weak nuclear force, emphasizing the vast difference in their magnitudes.

Given,

Hydrogen atom $Z=1$

1st excited state $n=2$

2nd excited state $n=3$

Energy associated with the $n$th orbit of an atom with $z$ atomic number

$ E=\frac{-13.6}{n^2}\left(Z^2\right) \mathrm{eV} $

Energy for 1st excited state

$ E_1=\frac{-13.6}{4} \mathrm{eV} $

Energy for 2nd excited state

$ E_2=\frac{-13.6}{9} \mathrm{eV} $

The ratio of $E_2 / E_1$ is $9 / 4$

$ \text { Hence, the ratio } E_2: E_1 \text { is } 9: 4 $

In the given nuclear reaction:

$ {}_{13} \mathrm{Al}^{27} + {}_2 \mathrm{He}^4 \rightarrow {}_0 n^1 + X $

We are asked to identify the element $ X $.

Conservation of Mass Number

According to the law of conservation of mass number, the total number of protons and neutrons (mass number) must be equal on both sides of the reaction:

$ 27 + 4 = 1 + a $

Solving for $ a $:

$ a = 30 $

This indicates that the mass number of element $ X $ is 30.

Conservation of Atomic Number

Similarly, the atomic number, which is the number of protons, must also be conserved:

$ 13 + 2 = 0 + b $

Solving for $ b $:

$ b = 15 $

This tells us that the atomic number of element $ X $ is 15.

Therefore, the element $ X $ with atomic number 15 and mass number 30 is:

$ {}_{15} \mathrm{P}^{30} $

Thus, the element $ X $ is phosphorous-30 ($ {}_{15} \mathrm{P}^{30} $).

Nuclear fission and fusion are explained using Einstein's mass-energy equation, $ E = \Delta m c^2 $.

Nuclear Fission:

This is the process where a heavy nucleus splits into two or more lighter nuclei.

Nuclear Fusion:

This occurs when two or more light nuclei combine to form a heavier nucleus.

In both processes, the mass difference between the reactants and the products ($\Delta m$) can be converted into energy, as described by the equation. Here, $\Delta m$ represents the difference in mass between the reactants and the products. The energy produced is directly proportional to this mass difference, multiplied by the square of the speed of light ($c^2$).

To understand the relative strengths of different fundamental forces, we need to compare their orders of magnitude. Here are the forces in question:

$F_1$: Gravitational Force

$F_2$: Weak Nuclear Force

$F_3$: Electromagnetic Force

The ratios depicting the relative strengths of these forces are as follows:

Gravitational Force: 1

Weak Nuclear Force: $10^{26}$

Electromagnetic Force: $10^{36}$

This means that the gravitational force is the weakest, while the electromagnetic force is the strongest among these three. Thus, the forces can be ordered by strength as:

$ F_1 < F_2 < F_3 $

Here, $ F_1 $ represents the gravitational force, $ F_2 $ the weak nuclear force, and $ F_3 $ the electromagnetic force.

To solve this, we use the Bohr quantization rule for angular momentum:

In the Bohr model, the angular momentum $L$ of an electron in a hydrogen atom is given by

$L = n\hbar,$

where $n$ is the principal quantum number and

$\hbar = \frac{h}{2\pi}$ is the reduced Planck's constant.

For the 4th orbit, we have $n = 4$. Plugging this into the formula gives:

$L = 4\hbar = 4 \left(\frac{h}{2\pi}\right) = \frac{4h}{2\pi} = \frac{2h}{\pi}.$

Thus, the angular momentum in the 4th orbit is $\frac{2h}{\pi}$.

So, the correct answer is Option B.

To find the energy equivalent of a 1 kg mass, we use Einstein's famous equation:

$ E = mc^2 $

Here's how to calculate it:

Given:

Mass, $m = 1\, \text{kg}$

Speed of light, $c \approx 3 \times 10^8 \, \text{m/s}$

Substitute the values into the equation:

$ E = 1 \times \left(3 \times 10^8\right)^2 $

Calculating the square of $3 \times 10^8$:

$ \left(3 \times 10^8\right)^2 = 9 \times 10^{16} $

Thus, the energy equivalent is:

$ E = 9 \times 10^{16} \, \text{J} $

Therefore, the correct answer is Option C: $9 \times 10^{16} \, \text{J}$.

The surface area of a nucleus with mass number $ A $ can be determined as follows:

Radius Relationship: The radius $ R $ of a nucleus is proportional to $ A^{1/3} $, given by the formula:

$ R = R_0 A^{1/3} $

Here, $ R_0 $ is a constant.

Surface Area Calculation: Since the nucleus is approximately spherical, its surface area $ S $ is calculated using the formula for the surface area of a sphere:

$ S = 4 \pi R^2 $

Substituting the expression for $ R $, we get:

$ S = 4 \pi \left(R_0 A^{1/3}\right)^2 $

Proportional Relation: Simplifying this expression, we find:

$ S \propto A^{2/3} $

Thus, the surface area $ S $ is proportional to $ A^{2/3} $.

We know,

Radius of nucleus, $r = {r_0}{A^{{1 \over 3}}}$

where, A = mass number

$\therefore$ Volume $(v) = {4 \over 3}\pi {r^3}$

$ = {4 \over 3}\pi r_0^3\,.\,A$

$\therefore$ $v \propto A$

$\therefore$ Volume is directly proportional to mass number.

We know,

density $(d) = {m \over v}$

$ = {{z{m_p} + (A - z){m_N}} \over {{4 \over 3}\pi r_0^3{{({A^{1/3}})}^3}}}$

$ = {{z{m_p} + A{m_p} - z{m_p}} \over {{4 \over 3}\pi r_0^3A}}$ [$\because$ ${m_p} \simeq {m_N}$]

$ = {{A{m_p}} \over {{4 \over 3}\pi r_0^3A}}$

$ = {{{m_p}} \over {{4 \over 3}\pi r_0^3}}$

$\therefore$ Density of nucleus is independent of mass number.

${E_1} = {E_0}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right) = {E_0} \times {3 \over 4}$

${E_2} = {E_0}$

$\therefore$ ${{{E_1}} \over {{E_2}}} = {3 \over 4}$

$N = {{{N_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$

$ \Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{N_0}} \over N} = {{{N_0}} \over {\left( {{{{N_0}} \over 8}} \right)}} = 8$

$ \Rightarrow {t \over {{T_{1/2}}}} = 3$

$ \Rightarrow {T_{1/2}} = {{15} \over 3} = 5$ min

$\because$ $N = {{{N_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$

$ \Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{N_0}} \over N} = {{{N_0}} \over {\left( {{{{N_0}} \over 8}} \right)}}$

$ \Rightarrow {2^{{t \over {{T_{1/2}}}}}} = 8 = {2^3}$

$ \Rightarrow t = 3 \times {T_{1/2}} = 3 \times 60$

$ = 180$ days

$\because$ $A = {{{A_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$

$ \Rightarrow {2^{t/5}} = {{6.4 \times {{10}^{ - 4}}} \over {5 \times {{10}^{ - 6}}}} = 128 = {2^7}$

$ \Rightarrow {t \over 5} = 7$

$ \Rightarrow t = 35$ days

$\because$ $A = {{{A_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$

$ \Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{A_0}} \over A} = 16$

$ \Rightarrow {t \over {{T_{1/2}}}} = 4$

$ \Rightarrow {{30} \over {{T_{1/2}}}} = 4$

$ \Rightarrow {T_{1/2}} = {{30} \over 4}$

$ = 7.5$ years

Linear momentum is conserved

so, ${p_{M'/3}} = {p_{2M'/3}}$

so, ${{{\lambda _{M'/3}}} \over {{\lambda _{2M'/3}}}} = {1 \over 1}$

$\therefore$ $R = {R_0}{A^{{1 \over 3}}}$

$ \Rightarrow {{{R_1}} \over {{R_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^{{1 \over 3}}} = {\left( {{4 \over 3}} \right)^{{1 \over 3}}}$

$\therefore$ Density ratio, ${{{\rho _1}} \over {{\rho _2}}} = {{{A_1}/{V_1}} \over {{A_2}/{V_2}}}$

$ = \left( {{{{A_1}} \over {{A_2}}}} \right) \times {\left( {{{{R_2}} \over {{R_1}}}} \right)^3}$

$ = 1:1$

${A_0} = 4250$

$A = 2250 = {A_0}{e^{ - \lambda t}}$

$ \Rightarrow {{2250} \over {4250}} = {e^{ - \lambda t}}$

$ \Rightarrow \lambda (10) = \ln \left( {{{4250} \over {2250}}} \right)$

$\lambda (10) = 0.636$

$\lambda = 0.063$

$\because$ ${1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$

$ \Rightarrow {1 \over {\lambda R}} = 1 - {1 \over {{n^2}}}$

$ \Rightarrow {1 \over {{n^2}}} = 1 - {1 \over {\lambda R}} = {{\lambda R - 1} \over {\lambda R}}$

$ \Rightarrow n = \sqrt {{{\lambda R} \over {\lambda R - 1}}} $

$\because$ $mvr = {{nh} \over {2\pi }}$

$ \Rightarrow mv = {{nh} \over {2\pi r}}$

$\because$ $\overrightarrow \mu = {{q\overrightarrow L } \over {2m}}$

$ \Rightarrow \overrightarrow \mu = {{ - e\overrightarrow L } \over {2m}}$

Equivalent reaction can be written as

$D\buildrel {} \over \longrightarrow {D_4} + 2\alpha + {\beta ^ - } + \gamma $

$\Rightarrow$ Mass number of D₄ = Mass number of D $-$ 2 $\times$ 4

= 182 $-$ 8 = 174

$\Rightarrow$ Atomic number of D₄

= Atomic number of D $-$ 2 $\times$ 2 + 1

= 74 $-$ 4 + 1 = 71

Radioactive decay is a random and spontaneous process it depends on unbalancing of nucleus.

$N = {N_0}{e^{ - \lambda t}}$ ..... (B)

$\ln N = - \lambda t + \ln {N_0}$

So, slope $ = - \lambda $ ..... (C)

${t_{1/2}} = {{\ln 2} \over \lambda }$

So ${t_{1/2}} \times \lambda = \ln 2 = $ Constant

K_p > 0

If Q is released $\Rightarrow$ Q > 0

$\Rightarrow$ K_p + Q > 0

Even the particle has to be given kinetic energy greater than magnitude of Q to maintain momentum conservation.

$\Rightarrow$ K + Q > 0

Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.

Also, E₂ $-$ E₁ is the energy of photon

$\Rightarrow$ E₂ $-$ E₁ = hf

$ \Rightarrow f = {{{E_2} - {E_1}} \over h}$

$ - 13.6 + 10.2 = {{ - 13.6} \over {{n^2}}}$

$ \Rightarrow {{13.6} \over {{n^2}}} = 3.4$

$ \Rightarrow n = 2$

$ \Rightarrow \Delta L = 2 \times {h \over {2\lambda }} - 1 \times {h \over {2\lambda }}$

$ = {h \over {2\lambda }}$

$ \Rightarrow \Delta L \simeq 1.05 \times {10^{ - 34}}$ Js