Atoms and Nuclei

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.