Atoms and Nuclei

$ \begin{aligned} & \frac{K Q^2}{2 r}=3.4 \mathrm{eV} \\ & \frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{2 \times 3.4 \times 1.6 \times 10^{-19}}=r \\ & 2.1176 \times 10^{-10}=r \\ & r=2.1 \times 10^{-10} \mathrm{~m} \end{aligned} $

As we know,

Size of nucleus, $r=r_0(A)^{\frac{1}{3}}$

$ \begin{aligned} & \Rightarrow \quad r^3=r_0^3 \cdot A \\ & \Rightarrow \quad V \propto A \quad\left(\because V=\frac{4}{3} \pi r^3\right) \end{aligned} $

So, option $\mathrm{A} \rightarrow$ wrong and $\mathrm{B} \rightarrow$ correct while the difference between the actual mass of nucleus and its constituents is called the mass defect.

∴ option $\mathrm{C} \rightarrow$ wrong and $\mathrm{D} \rightarrow$ correct

Given

$\rho=2.29 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3$, mass $m=19.926 \times 10^{-27} \mathrm{~kg}$

$R_0=1.2 \times 10^{-15} \mathrm{~m}$ and $R=R_0 A^{\frac{1}{3}}$

Now use volume $=\frac{\text { Mass }}{\text { Density }}$

$\Rightarrow \quad \frac{4}{3} \pi R^3=\frac{M}{\rho}$

$\Rightarrow \quad \frac{4}{3} \pi\left[R_0(A)^{1 / 3}\right]^3=\frac{M}{\rho}$

$\Rightarrow \quad \frac{4}{3} \pi R_0^3 A=\frac{M}{\rho}$

$\Rightarrow \quad A=\frac{M}{\rho} \times \frac{3}{4 \pi \times R_0^3}=\frac{19.926 \times 10^{-27} \times 3}{2.29 \times 10^{17} \times 12.56 \times\left(1.2 \times 10^{-15}\right)^3} \approx 12$

When a particle of mass $m$ is subject to a constant force $F$ pulling it toward the origin, we can apply the Bohr model to describe its motion.

Starting with the relationship for centripetal force:

$ F = \frac{m v^2}{r} $

We can rearrange this to show:

$ \frac{v^2}{r} = \text{constant} $

From this, it follows that:

$ r \propto v^2 \quad \text{...(1)} $

In the Bohr model, the angular momentum $L$ is quantized and given by:

$ L = m v r = \frac{n h}{2 \pi} \quad \text{...(2)} $

Solving equations (1) and (2) together, we find:

$ v \propto n^{1/3} $

Substituting this into equation (1), we get:

$ r \propto n^{2/3} $

Thus, the radius of the $n^{\text{th}}$ orbit and the speed of the particle depend on $n$ as follows:

$r \propto n^{2/3}$

$v \propto n^{1/3}$

The spectral series corresponds to groups of wavelengths that are produced when electrons in an atom make transitions between energy levels. Each series is named after the scientist who discovered it and is characterized by the energy level to which the electrons transition.

The transitions to the level $n_1 = 4$ define the Brackett series. Electrons that fall to the fourth energy level (from higher levels with $n_2 = 5, 6, \ldots$) emit or absorb radiation in the infrared region of the electromagnetic spectrum.

Here's a quick summary of the different spectral series:

• Pfund series: Transitions to $n_1 = 5$
• Brackett series: Transitions to $n_1 = 4$
• Lyman series: Transitions to $n_1 = 1$
• Balmer series: Transitions to $n_1 = 2$

Therefore, the correct answer is:

Option B: Brackett series

Water is used as a coolant in a nuclear reactor primarily because of its high specific heat capacity. High specific heat capacity means that water can absorb a large amount of heat energy with only a small increase in its own temperature. This property makes water an effective medium for transferring heat away from the reactor core, thereby maintaining safe operational temperatures and preventing overheating.

Therefore, the correct answer is:

Option B: high specific heat capacity

$\begin{aligned} & h \frac{c}{\lambda_1}=\frac{-5 E}{2}+4 E=\frac{3}{2} E \quad \text{.... (1)}\\ & h \frac{c}{\lambda_2}=-2 E+3 E=E \quad \text{.... (2)}\\ & h \frac{c}{\lambda_3}=-2 E+4 E=2 E \quad \text{.... (3)} \end{aligned}$

Comparing (2) and (3)

$\frac{1}{\lambda_3}=\frac{2}{\lambda_2} \quad \lambda_2=2 \lambda_3$

Comparing (1) and (2)

$3 \lambda_1=2 \lambda_2 \quad \lambda_1<\lambda_2$

(A) Slow neutron can cause fission in ${ }_{92}^{235} \mathrm{U}$ than fast neutrons because fast neutrons are too quick so they scatter of atoms instead of being captured by them

(B) $\alpha$-rays are Helium nuclei, is a true statement

(C) $\beta$-rays are generated when neutron is converted into proton by releasing electron or proton is converted into neutron by releasing positron.

(D) $\gamma$-rays have higher energies as compared to X-rays. So they have smaller wavelength as compared to X-rays.

Let's analyze each statement one by one to determine the correctness:

Statement I: "Atoms are electrically neutral as they contain equal number of positive and negative charges."

This statement is correct. Atoms are composed of protons, neutrons, and electrons. Protons have a positive charge, electrons have a negative charge, and neutrons have no charge. In a neutral atom, the number of protons (positive charges) equals the number of electrons (negative charges), thus making the atom electrically neutral. The positive and negative charges balance each other, resulting in no overall charge.

Statement II: "Atoms of each element are stable and emit their characteristic spectrum."

This statement requires modification for accuracy. Atoms emit their characteristic spectrum when they are excited and then return to a lower energy state. The emission of a spectrum is not linked directly to the stability of an atom; it is linked to the electronic transitions within the atom. The term "stable" in relation to atoms typically refers to the overall energy state of an atom being at its lowest or in a ground state. However, atoms can emit radiation whether they are in a stable ground state or in an excited state. Moreover, not all atoms are inherently stable; some may be radioactive and unstable. The part that atoms emit their characteristic spectrum under certain conditions (like excitation) is correct, but linking this property strictly to stability is misleading.

Based on the analysis:

• Statement I is correct.
• Statement II is partially correct, as it is accurate that each element can emit a characteristic spectrum when excited, but incorrect in generalizing that all such atoms are stable.

Therefore, the correct option is:

Option C: Statement I is correct but Statement II is incorrect.

$\begin{aligned} & \text { Energy difference } \Delta E=\frac{h c}{\lambda} \\ & \therefore \lambda \propto \frac{1}{\Delta E} \\ & (\Delta E)_{6-2}>(\Delta E)_{5-2}>(\Delta E)_{4-2}>(\Delta E)_{3-2} \\ & \lambda_{6-2}<\lambda_{5-2}<\lambda_{4-2}<\lambda_{3-2} \end{aligned}$

A-III, B-IV, C-II, D-I

To determine the mass number and atomic number of the final product $Q$ in the provided nuclear reactions, we need to analyze the impact of each type of decay (alpha, positron emission, beta-minus decay, and electron capture) on the mass number and atomic number of the initial element $X = { }_{82}^{290} X$:

1. Alpha decay ($\alpha$ decay): In alpha decay, an alpha particle (which is a $^4_2He$ nucleus) is emitted. This reduces the mass number by 4 units and the atomic number by 2 units.

$ { }_{82}^{290} X \xrightarrow{\alpha} { }_{80}^{286} Y $

2. Beta plus decay (positron emission, $e^{+}$): During positron emission, a proton in the nucleus is transformed into a neutron, and a positron is emitted. This decreases the atomic number by 1 but does not change the mass number.

$ { }_{80}^{286} Y \xrightarrow{e^{+}} { }_{79}^{286} Z $

3. Beta-minus decay ($\beta^{-}$ decay): In a beta-minus decay, a neutron in the nucleus converts into a proton and an electron (beta particle) and an antineutrino are emitted. This results in an increase in the atomic number by 1, while the mass number remains unchanged.

$ { }_{79}^{286} Z \xrightarrow{\beta^{-}} { }_{80}^{286} P $

4. Electron capture ($e^{-}$ capture): During electron capture, an atomic electron is absorbed by the nucleus, causing a proton to convert into a neutron. This process decreases the atomic number by 1 without altering the mass number.

$ { }_{80}^{286} P \xrightarrow{e^{-}} { }_{79}^{286} Q $

From the calculations above, the mass number of $Q$ is 286, and its atomic number is 79. Comparing these values with the multiple choices:

Option A: Mass number = 280, Atomic number = 81
Option B: Mass number = 286, Atomic number = 80
Option C: Mass number = 288, Atomic number = 82
Option D: Mass number = 286, Atomic number = 81

None of these match exactly, implying an error in the problem or the answer options. Based on the correct atomic number calculations and assuming option D is meant to be atomic number 79, then the corrected choice would have been:

D: Mass number = 286, Atomic number = 79

The energy levels of a hydrogen atom are given by the formula :

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

where $E_n$ is the energy of the $n$-th energy level.

The ground state of hydrogen ($n=1$) has an energy of $-13.6 ~\mathrm{eV}$ as mentioned, which means that it would take $+13.6 ~\mathrm{eV}$ to ionize it (remove the electron completely) from this state, since ionization implies moving the electron to a state of zero energy.

The second excited state of hydrogen is when $n=3$ (as $n=1$ is the ground state and $n=2$ is the first excited state). Thus, the energy of the second excited state is :

$E_3 = -\frac{13.6 ~\mathrm{eV}}{3^2} = -\frac{13.6 ~\mathrm{eV}}{9} = -1.51 ~\mathrm{eV}$

Since ionization implies moving the electron from its current energy level to $0$ energy, the energy required to ionize the atom from this state is the absolute value of its current energy state. So, it will take $+1.51 ~\mathrm{eV}$ to ionize a hydrogen atom from its second excited state.

So, the correct answer is Option C : $1.51 ~\mathrm{eV}$.

$ \begin{aligned} & \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{(1)^2}-\frac{1}{\mathrm{n}^2}\right) \mathrm{n}=2,3,4, \ldots \ldots \\ & \left(\frac{1}{\lambda_{\mathrm{L}}}\right)_{\max }=\mathrm{R}\left(\frac{1}{(1)^2}-\frac{1}{(2)^2}\right) \quad\left(\because \frac{1}{\mathrm{R}} \simeq 912 \mathop A\limits^o\right) \\ & \left(\lambda_{\mathrm{L}}\right)_{\max }=\frac{4}{3} \frac{\mathrm{L}}{\mathrm{R}} \\ & \left(\lambda_{\mathrm{L}}\right)_{\max }=\frac{4}{3} \times 912 \mathop A\limits^o=4 \times 304 \mathop A\limits^o=1216 \mathop A\limits^o \\ & \left(\frac{1}{\lambda_{\mathrm{L}}}\right)_{\min }=\mathrm{R}\left(\frac{1}{(1)^2}-\frac{1}{(\infty)^2}\right) \\ & \left(\lambda_{\mathrm{L}}\right)_{\min }=\frac{1}{\mathrm{R}} \simeq 912 \mathop A\limits^o \end{aligned}$

Range of $\lambda$ is $912 \mathop A\limits^o$ to $1216 \mathop A\limits^o$ which lies in U.V. region.

The given angular momentum of the electron is $1.5\left(\frac{h}{\pi}\right)$, where $h$ is Planck's constant.

According to the Bohr model of the hydrogen atom, the allowed angular momenta for an electron are quantized and given by $mvr = n \frac{h}{2\pi}$ where $m$ is the mass of the electron, $v$ is its velocity, $r$ is the radius of the orbit, and $n$ is the principal quantum number (an integer).

By comparing the given angular momentum with the quantized form, we get $1.5 \frac{h}{\pi} = n \frac{h}{2\pi}$, which simplifies to $n = 3$.

Now, we use the formula for the energy levels of the hydrogen atom: $E_n = -\frac{13.6 \text{eV}}{n^2}$. Since $n = 3$, the energy can be calculated as $E_3 = -\frac{13.6 \text{eV}}{3^2}$, which is approximately $-1.5 \text{eV}$.

Half life $T=20 \mathrm{~min}$

Left fraction of activity $\frac{1}{16}$

$ \begin{aligned} \because \frac{\mathrm{R}}{\mathrm{R}_{0}} & =\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}} \\ \frac{1}{16} & =\left(\frac{1}{2}\right)^{\mathrm{t} / 20} \\ \left(\frac{1}{2}\right)^{4} & =\left(\frac{1}{2}\right)^{\mathrm{t} / 20} \\ 4 & =\frac{\mathrm{t}}{20} \\ \mathrm{t} & =80 \mathrm{~min} \end{aligned} $

Shortest wavelength in Balmer series when transition of $e^{-}$ from $\infty$ to $\mathrm{n}=2$

$\because \frac{1}{\lambda}=\mathrm{Rz}^{2}\left[\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right]$

$\frac{1}{\lambda}=\frac{R}{4}$ .... (1)

Shortest wavelength is Bracket series when transition of $e^{-}$ from $\infty$ to $\mathrm{n}=4$

$\frac{1}{\lambda^{\prime}}=\mathrm{R}(1)^{2}\left[\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right] \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{\mathrm{R}}{16}$ ..... (2)

Eq. (1) / Eq. (2)

$\frac{\lambda^{\prime}}{\lambda}=\frac{\mathrm{R}}{4} \times \frac{16}{\mathrm{R}} \Rightarrow \lambda^{\prime}=4 \lambda$

The formula to calculate the radius of the nth orbit in a hydrogen atom is given by the equation:

$r_n = r_0 \times n^2$

where,
$r_n$ is the radius of the nth orbit,
$r_0$ is the radius of the innermost orbit (known as the Bohr radius), and
$n$ is the orbit number.

Given that the radius of the innermost orbit ($r_0$) is $5.3 \times 10^{-11} \mathrm{~m}$ or equivalently $0.53 \mathop A\limits^o$, and we're interested in finding the radius of the third orbit ($n = 3$).

So,

$r_3 = 0.53 \times (3)^2$ $\mathop A\limits^o$

$r_3 = 0.53 \times 9$ $\mathop A\limits^o$

$r_3 = 4.77$ $\mathop A\limits^o$

Therefore, the radius of the third allowed orbit of the hydrogen atom is 4.77 $\mathop A\limits^o$.

Radius of Bohr's orbit depends on principal quantum number (n) as

$R \propto {n^2}$

Now, ${{{R_1}} \over {{R_2}}} = {{{{(2)}^2}} \over {{{(4)}^2}}} = {1 \over 4} = 0.25$

According to law of radioactive decay

Rate of decay $\left( { - {{dN} \over {dt}}} \right) \propto N$ (No. of undecayed nuclei)

Half life is the time in which half of the radioactive sample decays

$\therefore$ Both statements are incorrect.

Number of radioactive nuclei at any time is

$N = {N_0}{e^{ - \lambda t}}$

Initial number of nuclei is same for sample X₁ & X₂.

Let after time 't' the ratio of their atoms become ${1 \over e}$

$ \Rightarrow {{{N_0}{e^{ - 10\lambda \times t}}} \over {{N_0}{e^{ - \lambda \times t}}}} = {1 \over e} \Rightarrow {e^{ - 9\lambda t}} = {e^{ - 1}} \Rightarrow t = {1 \over {9\lambda }}$

Hence, $t = {1 \over {9\lambda }}$

Ratio of magnitude of Coulomb's electrostatic force to the gravitational force

${{{F_E}} \over {{F_G}}} = {{\left( {{{K{q_1}{q_2}} \over {{r^2}}}} \right)} \over {\left( {{{G{m_1}{m_2}} \over {{r^2}}}} \right)}} = {{K{q_1}{q_2}} \over {G{m_1}{m_2}}}$

$ \Rightarrow 2.4 \times {10^{39}} = {K \over G} \times {{1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}} \over {9.11 \times {{10}^{ - 31}} \times 1.67 \times {{10}^{ - 27}}}}$

$ \Rightarrow 2.4 \times {10^{39}} = {K \over G} \times {{2.56} \over {15.21}} \times {10^{20}} \Rightarrow {K \over G} = 14.26 \times {10^{19}}$

$ \Rightarrow {K \over G} = 1.426 \times {10^{20}}$ $\Rightarrow$ Ratio $\approx$ 10²⁰

de-Broglie wavelength associated with a particle is given by

$\lambda = {h \over p}$

$\lambda \propto {1 \over p}$

The nuclear reaction is given as

${}_{11}^{22}Na \to {}_Z^AX + {}_{ + 1}{e^0} + v$

From conservation of atomic number

$11 = Z + 1 \Rightarrow Z = 10 \Rightarrow Ne$

From conservation of mass number

$22 = A + 0 \Rightarrow A = 22$

$\therefore$ ${}_Z^AX = {}_{10}^{22}Ne$

${E_n} = {{{E_0}} \over {{n^2}}}$, For first excited state $\Rightarrow$ n = 2

For second excited state $\Rightarrow$ n = 3

$ \Rightarrow {{{T_1}} \over {{T_2}}} = {{{{{E_0}} \over 4}} \over {{{{E_0}} \over 9}}} = {9 \over 4}$

Radius of nuclei with mass number A varies as

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

${{{R_1}} \over {{R_2}}} = {\left( {{{125} \over {64}}} \right)^{1/3}} = {5 \over 4} = 5:4$

Given, total energy given to $e^{-}$ in 4th orbit, $E=15 \mathrm{~eV}$

Energy of $n$th orbit of $\mathrm{H}$-atom, $E^{\prime}=-\frac{13.6}{n^2}$

So, energy of 4th orbit of $\mathrm{H}$-atom, $E^{\prime}$

$=\frac{-13.6}{4^2}=-0.85 \mathrm{~eV}$

$\therefore$ Ionisation energy of 4th orbit $=-E^{\prime}$

$=-(-0.85 \mathrm{eV})=0.85 \mathrm{~eV}$

$\therefore$ Final energy of electron, when it comes out of $\mathrm{H}$-atom.

$E=$ Total energy given to electron - Ionisation energy of 4th orbit

$=15-0.85=14.15 \mathrm{~eV}$

Given, half life of an element, $t_{1 / 2}=69.3 \mathrm{~h}$

Let initial amount (at $t=0$) of radioactive element $\left(t_{1 / 2}\right)=69.3 \mathrm{~h}$

Active nuclei at $t=10 \mathrm{~h}$,

$N_1=N_0 e^{-10 \lambda}$

Active nuclei at $t=11 \mathrm{~h}, N_2=N_0 e^{-11 \lambda}$

% decay in 10th to 11th $\mathrm{h}$

$\begin{aligned} & =\frac{N_1-N_2}{N_1} \times 100=\frac{N_0 e^{-10 \lambda}-N_0 e^{-11 \lambda}}{N_0 e^{-10 \lambda}} \times 100 \\ & =\left(1-e^{-\lambda}\right) \times 100 \\ & \begin{aligned} & =\left[1-e^{\frac{-0.693}{69.3}}\right] \times 100 \quad\left[\because \text { where, } \lambda=\frac{0.693}{t_{1 / 2}}\right] \\ & =\left[1-e^{-0.01}\right] \times 100=[1-0.99] \times 100 \\ & =0.01 \times 100=1 \% \end{aligned} \end{aligned}$

Heavy water $\left(\mathrm{D}_2 \mathrm{O}\right)$ is used to slow down fast moving neutrons in nuclear reactor, hence it is called moderator. It does remove neutrons from the system by absorbing them. When fast moving neutrons are passed through heavy water $\left(\mathrm{D}_2 \mathrm{O}\right)$, then they make elastic collisions with protons, which have comparatively much smaller than velocities. In few interactions, the velocities of neutrons get interchanged with those of protons. Heavy water does not react with slow moving neutrons, i.e. it slow down-fast moving neutrons. Hence, assertion and reason both are true and reason is the correct explanation of assertion.

For an element, it is not always true that number of neutrons $(N)$ are greater than or equal to atomic number $(Z)$.

For example, in hydrogen atom, there is no neutron but it has one proton.

Therefore, atomic number of hydrogen atom is greater than number of neutrons.

i.e., $Z>N$ for hydrogen atom.

The force between neutrons in the nucleus is attractive only upto the range of $1 \mathrm{~fm}$, But this attractive force become highly repulsive when neutrons came nearer to about 0.6 to $0.7 \mathrm{~fm}$. Therefore, neutrons do not experience always attractive nuclear force.

Hence, both assertion and reason are false.

$\begin{aligned} & \text { } \therefore N_1=N_0-\frac{1}{3} N_0=\frac{2}{3} N_0 \\ & \text { and } \quad N_2=N_0-\frac{2}{3} N_0=\frac{1}{3} N_0 \\ & \therefore \quad \frac{N_2}{N_1}=\left(\frac{1}{2}\right)^n \Rightarrow n=1 \\ & \therefore t_2-t_1=\text { one half-life }=20 \mathrm{~min} \end{aligned}$

In an atom electron revolves around nucleus, for this required centripetal force is provided by electrostatic force of attraction between electron and nucleus. As there is no other force, electrostatic force is unbalanced.

For a $\beta^{-}$-decay,

${ }_0^1 n \longrightarrow{ }_1^1 p+{ }_{-1}^0 e$

$\therefore$ Electrons do not exist inside the nucleus.

Energy released per fission is

$\begin{aligned} E & =200 \mathrm{MeV} \\ & =200 \times 1.6 \times 10^{-13} \\ & =3.2 \times 10^{-11} \mathrm{~J} \end{aligned}$

Total energy required to run a $1 \mathrm{~MW}$ reactor for one year is

$\begin{aligned} & =10^6 \mathrm{Js}^{-1} \times(365 \times 24 \times 60 \times 60) \\ & =3.15 \times 10^{13} \mathrm{~J} \end{aligned}$

Since, 1 fission ( 1 atom of $\mathrm{U}^{235}$ ) produces $3 \cdot 2 \times 10^{-11} \mathrm{~J}$ of energy. So, total number of $\mathrm{U}^{235}$ atoms required is

$\frac{3.15 \times 10^{13}}{3.2 \times 10^{-11}}=9.84 \times 10^{23}$

Now, $6.02 \times 10^{23}$ atoms of $\mathrm{U}^{235}$ are contained in $235 \mathrm{~g}$ of $\mathrm{U}^{235}$

$\therefore$ Mass of $\mathrm{U}^{235}$ containing $9.84 \times 10^{23}$ atoms is

$\begin{aligned} M & =\frac{235}{6.02 \times 10^{23}} \times 9.84 \times 10^{23} \\ & =384 \mathrm{~g}=0.384 \mathrm{~kg} \end{aligned}$

Option A: Both assertion and reason are true and reason is the correct explanation of assertion.

Using a beam of charged particles, such as protons or heavy-ion beams, is an effective method in cancer treatment. This technique is known as particle therapy or proton therapy. The reason for employing charged particles in cancer treatment lies in their ability to deposit energy specifically at the tumor site through ionization, causing minimal damage to surrounding healthy tissue.

When charged particles pass through a material medium, they interact with the atoms of the medium, primarily by ionizing them. During this process, they lose energy, which is deposited along their path. This ability to precisely control the deposition of energy at a specific location (the tumor) with a sharp "Bragg peak" at the end of their range is what makes them particularly useful in treating cancerous tissues effectively while reducing potential side effects.

The transition in He-Ne Laser is shown below.

An electric discharge is the gas pump the helium atom to higher energy states. These He atoms collide with the ground state atoms and excite to higher states and produce population inversion causing a directional photon beam.

We know that $\alpha$-particle carries two units of positive charge and four units of mass. In $\alpha$-decays, charge number of parent nucleus decreases by 2 units. Also mass number reduces by four units.