Atoms and Nuclei

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} $.

Order of magnitude of fundamental forces is Gravity ( $10^{-38}$ ); weak nuclear Forces $\left(10^{-13}\right)$; electromagnetic $\left(10^{-2}\right)$; strong nuclear force $\left(10^0\right)$.

∴ Strong nuclear forces are with largest magnitude.

Wavelength of spectral lines of $n$-atom is given as

$ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

For shortest wavelength of H - atom spectrum in Balmer series,

$ \begin{aligned} n_1=2 \text { and } & n_2=\infty \\ \therefore \quad \frac{1}{\lambda} & =R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right) \\ \frac{1}{\lambda} & =\frac{R}{4} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \lambda & =\frac{4}{R} \\ & =\frac{4}{1.097 \times 10^7}=3.646 \times 10^{-7} \mathrm{~m} \\ & =3646 \times 10^{-10} \mathrm{~m}=3646 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

Total energy released = Total number of fission × Energy released per fission

$ \begin{aligned} & =m \times \frac{N}{M} \times 200 \times 10^6 \times 1.6 \times 10^{-19} \\ & =1 \times 10^{-3} \times \frac{6 \times 10^{23}}{240} \\ & \quad \times 200 \times 10^6 \times 1.6 \times 10^{-19} \\ & =8 \times 10^7 \mathrm{~J} \end{aligned} $

Range of fundamental forces of nature (in ascending order) is

Weak nuclear force (less than $10^{-18} \mathrm{~m}$ );

Strong nuclear force (less than $10^{-15} \mathrm{~m}$ );

Electromagnetic force (very large) ;

Gravity (infinity)

Strength of forces (in ascending order) is

Gravity ( $10^{-38}$ ) :

weak nuclear ( $10^{-13}$ );

electromagnetic $\left(10^{-2}\right)$;

strong nuclear force $\left(10^{\circ}\right)$.

The energy of an electron in $n$th excited state of H -atom is given as

$ E_n=\frac{-13.6}{n^2} \mathrm{eV} $

For fourth excited state,

$ n=5 $

$ \begin{aligned} \therefore \quad E_5 & =\frac{-13.6}{5^2} \\ & =\frac{-13.6}{25} \\ & =-0.54 \mathrm{eV} \end{aligned} $

Given, for Al nucleus

Mass number, $A=27$

$ \begin{aligned} R_0 & =1 \times 10^{-15} \mathrm{~m} \\ \pi & =3 \end{aligned} $

and

We know that, radius of nucleus is given as

$ \begin{aligned} R & =R_0 A^{1 / 3} \\ & =1 \times 10^{-15} \times(27)^{1 / 3} \\ & =3 \times 10^{-15} \mathrm{~m} \end{aligned} $

Volume of aluminium (Al) nucleus,

$ \begin{aligned} V & =\frac{4}{3} \pi R^3 \\ & =\frac{4}{3} \times \pi \times\left(3 \times 10^{-15}\right)^3 \\ & =\frac{4}{3} \times 3 \times 27 \times 10^{-45} \\ & =108 \times 10^{-45} \mathrm{~m}^3 \\ & =1.08 \times 10^{-13} \times\left(10^{-10}\right)^3 \mathrm{~m}^3 \\ & =1.08 \times 10^{-13} \times(1 \mathop {\rm{A}}\limits^{\rm{o}})^3 \\ & \simeq 1 \times 10^{-13} {\mathop {\rm{A^3}}\limits^{\rm{o}} } \end{aligned} $

The nuclear force is a powerful attractive force between nucleons at a distance of about 1 femtometer or $1 \times 10^{-15} \mathrm{~m}$.

∴ Hence, range of nuclear force

$ =10^{-15} \mathrm{~m} $

According to Bohr's model of hydrogen atom, velocity of electron in $n$th orbit is given as

$ v \propto \frac{1}{n} $

For fourth orbit, $n_4=4$

For 9 th orbit, $n_9=9$

$ \begin{array}{rlrl} \therefore & \frac{v_4}{v_9} & =\frac{n_9}{n_4}=\frac{9}{4} \\ \therefore & v_4 & : v_9=9: 4 \end{array} $

Radius of ${ }^{189} \mathrm{O}$,

$ R=R_0(189)^{1 / 3} \quad\left[\because R=R_0 A^{1 / 3}\right] $

∴ Radius of given nucleus,

$ \begin{aligned} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, R^{\prime} =\frac{1}{3} R \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, R^{\prime} =\frac{1}{3} R_0(189)^{1 / 3} \\ &But \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,R^{\prime} =R_0\left(A^{\prime}\right)^{1 / 3} \\ & \Rightarrow \frac{1}{3} R_0(189)^{1 / 3} = R_0\left(A^{\prime}\right)^{1 / 3} \\ & \Rightarrow \quad \frac{1}{3}(189)^{1 / 3} = \left(A^{\prime}\right)^{1 / 3} \\ & \Rightarrow \quad\left(A^{\prime}\right)^{1 / 3} = \frac{1}{3}(189)^{1 / 3} \end{aligned} $

Cubing both side, we get

$ A^{\prime}=\frac{1}{3^3} \cdot 189=\frac{189}{27}=7 $

We know that, wavelength of spectral line in Balmer series is given as,

$ \frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

where $n_1=2$ and $n_2=3,4 \,\,\,\ldots.(i)$

$ \Rightarrow \quad \frac{1}{\lambda}=R_H\left[\frac{1}{2^2}-\frac{1}{n_2^2}\right] $

For maximum wavelength $n_2=3$

Form Eq. (i), we get

$ \begin{aligned} \frac{1}{\lambda_{\max }} & =R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ & =\frac{5}{36} R_H \\ \Rightarrow \quad \lambda_{\max } & =\frac{36}{5 R_H} \\ & =\frac{36}{5 \times 1 \times 10^7}=7.2 \times 10^{-7} \mathrm{~m} \\ \Rightarrow \quad \lambda_{\max } & =7200 \times 10^{-10} \mathrm{~m} \\ & =7200\mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

For minimum wavelength,

$ n_2=\infty $

$ \begin{aligned} \frac{1}{\lambda_{\min }} & =R_H\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=\frac{R_H}{4} \\ \Rightarrow \quad \lambda_{\min } & =\frac{4}{R_H}=\frac{4}{1 \times 10^7} \end{aligned} $

$ \begin{aligned} & =4 \times 10^{-7} \mathrm{~m} \\ \Rightarrow \quad \lambda_{\min } & =4000 \times 10^{-10} \mathrm{~m}=4000 \mathop {\rm{A}}\limits^{\rm{o}} \\ \therefore \lambda_{\max }-\lambda_{\min } & =7200-4000=3200\mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

We know that, radius of nucleus in terms of mass number $(A)$ is given as

$ \begin{aligned} R & =R_0\left(A^{1 / 3}\right) \\ \Rightarrow \quad \frac{R_2}{R_1} & =\left(\frac{A_2}{A_1}\right)^{\frac{1}{3}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(i)\end{aligned} $

But, $A_2=A_1+2 \%$ of $A_1$

$ =A_1+\frac{2}{100} A_1=\frac{102}{100} A_1=1.02 A_1 $

From Eq (i), we get

$ \begin{aligned} \frac{R_2}{R_1} & =\left(\frac{1.02 A_1}{A_1}\right)^{\frac{1}{3}}=(1.02)^{1 / 3} \\ R_2 & =R_1(1.02)^{1 / 3} \\ R_2 & =1.0067 R_1=R_1+0.0067 R_1 \\ & =R_1+0.67 \% \text { of } R_1 \\ & =R_1+\frac{2}{3} \% \text { of } R_1 \end{aligned} $

Hence, $R_2$ is larger than $R_1$ by $\frac{2}{3} \%$.

Weak nuclear force is responsible for radioactive decays and hence these forces are also responsible for beta decay.

We know that, frequency of spectral series of hydrogen atom is given as

$ v=R C \quad\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] $

For series limit of Balmer series,

$ \begin{aligned} & & n_1 & =2 \text { and } n_2=\infty \\ \therefore & & v_B & =R C\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right) \\ \Rightarrow & & v_B & =\frac{R C}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

For the series limit of Brackett series,

$ n_1=4 \text { and } n_2=\infty $

$ \therefore \quad v_B^{\prime}=R C\left(\frac{1}{4^2}-\frac{1}{\infty^2}\right)=\frac{R C}{16} $

Given, I amu $=1.6 \times 10^{-27} \mathrm{~kg}$

Since, nuclear density is independent of mass number, the nuclear density of nucleus ${ }_{30}^{60} \mathrm{X}$ is given as :

$ \rho=\frac{3 m}{4 \pi R_0^3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $m$ is average mass of nucleon.

$ \begin{aligned} & \text { Average mass of nucleon }=\frac{m_n+m_p}{2} \\ & \qquad \begin{aligned} & =\frac{1.0073+1.0086}{2} \mathrm{amu} \\ & =\frac{2.0159}{2}=1.00795 \mathrm{amu} \\ & =1.00795 \times 1.6 \times 10^{-27} \mathrm{~kg} \\ \Rightarrow \quad m & =1.61272 \times 10^{-27} \mathrm{~kg} \end{aligned} \end{aligned} $

∴ Putting the values in Eq. (i), we get

$ \begin{aligned} \rho & =\frac{3 \times 1.61272 \times 10^{-27}}{4 \times 314 \times\left(1.2 \times 10^{-15}\right)^3} \\ & =2.2 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3} \end{aligned} $

Let $A$ be the mass number and $R$ be the radius of a nucleus. If $m$ be the average mass of a nucleon, then mass of nucleus, $M=m A$

Volume of nucleus, $V=\frac{4}{3} \pi R^3$

∴ Nuclear density,

$ \begin{aligned} \rho & =\frac{\text { Mass of nucleus }(M)}{\text { Volume }(V)} \\ & =\frac{m A}{\frac{4}{3} \pi R^3}=\frac{3 m A}{4 \pi R^3} \\ & =\frac{3 m A}{4 \pi\left(R_0 A^{\frac{1}{3}}\right)^3} \quad\left(\because R=R_0 A^{1 / 3}\right) \\ & =\frac{3 m A}{4 \pi R_0^3 A}=\frac{3 m}{4 \pi R_0^3} \end{aligned} $

Hence, nuclear density does not depend on mass number.

Therefore, with increase of mass number A, nuclear density does not change.

Given, binding energy (BE) per

nucleon $=7.14 \mathrm{MeV}$

BE of element $=28.6 \mathrm{MeV}$

We know that,

$ \begin{aligned} \text { Number of nucleons } & =\frac{\text { BE of element }}{\text { BE per nucleon }} \\ & =\frac{28.6}{7.14}=4 \end{aligned} $

Wavelength for Lyman series is calculated as

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

For first line of Lyman series, $n=2$

$ \begin{array}{ll} \therefore & \frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \Rightarrow \frac{1}{\lambda}=\frac{3 R}{4} \\ \Rightarrow & \lambda=\frac{4}{3 R} \Rightarrow R=\frac{4}{3 \lambda} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

Wavelength of Balmer series is calculated as

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

For first line of Balmer series, $n=3$

$ \begin{array}{ll} \therefore & \frac{1}{\lambda^{\prime}}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ \Rightarrow & \frac{1}{\lambda^{\prime}}=\frac{5}{36} R \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{5}{36} \times \frac{4}{3 \lambda}\,\,\,\,\,\text { [from Eq. (i)] } \\ \Rightarrow & \frac{1}{\lambda^{\prime}}=\frac{5}{27 \lambda} \Rightarrow \lambda^{\prime}=\frac{27 \lambda}{5} \end{array} $

Half-life of radioactive isotope,

$ T_{1 / 2}=30 \mathrm{~h} $

If $N_0$ be the initial amount of radioactive isotope, then remaining amount $(N)$ in time $t$ is given as

$ N=125 \% \text { of } N_0=\frac{125}{100} \times N_0 \Rightarrow N=\frac{N_0}{8} $

We know that,

$ \begin{aligned} \quad N & =N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{8}=N_0\left(\frac{1}{2}\right)^n \\ \Rightarrow \quad\left(\frac{1}{2}\right)^3 & =\left(\frac{1}{2}\right)^n \Rightarrow n=3 \Rightarrow \frac{t}{T_{1 / 2}}=3 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad t & =3 \times T_{1 / 2}=3 \times 30=90 \mathrm{~h} \end{aligned} $

In atomic scale, the order of magnitude of forces in nature are related as,

Strong nuclear force> electromagnetic force > weak nuclear force > gravitational force

Hence, gravitational force is weakest force in nature in atomic scale.

In spectral line series, wavelength is given by

$ \frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) $

For Balmer transitions, $n_f=2$

$ \Rightarrow \quad \frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{n^2}\right) $

$ \begin{aligned} & \text { Given, } \lambda=16 / 3 R \\ & \Rightarrow \frac{3 R}{16}=R\left(\frac{1}{4}-\frac{1}{n^2}\right) \Rightarrow \frac{3}{16}=\frac{1}{4}-\frac{1}{n^2} \\ & \Rightarrow \frac{1}{n^2}=\frac{1}{4}-\frac{3}{16}=\frac{1}{16} \Rightarrow n=4 \end{aligned} $

Decay scheme of given sample is as shown below

So, time required by sample to reduced to 7.5 g is

$ \Delta t=3 \times T_{1 / 2}=3 \times 5=15 \mathrm{~s} $

Nuclear forces are effective within the nucleus (radius $\approx 10^{-15} \mathrm{~m}$ or less) and these are responsible for keeping protons within this small volume. Hence, nuclear forces are short range attractive in nature.

Wavelength of spectral lines of Hydrogen like atom is given as

$ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

For Balmer series, $n_1=2$ and $n_2=3,4,5,6, \ldots$.

$ \therefore \quad \frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For minimum wavelength, $n_2=\infty$ From Eq. (i),

$ \begin{aligned} \frac{1}{\lambda_{\min }} & =R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right) \\ \frac{1}{\lambda_{\min }} & =\frac{R}{4} \\ \Rightarrow \quad \lambda_{\min } & =\frac{4}{R} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

For maximum wavelength, $n_2=3$

∴ From Eq. (i), we have

$ \begin{aligned} \frac{1}{\lambda_{\max }} & =R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac{5}{36}\right) \\ \Rightarrow \lambda_{\max } & =\frac{36}{5 R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

∴ From Eq. (ii) and (iii), we have

$ \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{36 / 5 R}{4 / R}=\frac{36}{20}=\frac{9}{5} $

$\alpha$-rays are a bunch of particles consisting of 2 neutrons and 2 protons. Hence, a $\alpha$-particle is a doubly ionised helium-atom.