Structure of Atom

O^2- , F^- , Na⁺ and Mg²⁺, all have 10 electrons each.

- Option (A) : Incorrect. The total number of radial nodes is given by $n-l-1$. For $1 s$ orbital, the radial node is zero (i.e. $1-0-1=0$ ).

- Option (B) : Incorrect. $d_{z^2}$ orbital has two lobes that point in opposite directions along $z$-axis plus a doughnut-shaped ring of electron density around the centre that lies in the $x y$-plane. Therefore, it does not have any nodal plane.

- Option (C) : Correct. The total number of radial nodes is given by $n-l-1$.

For $2 s$ : Radial nodes $=2-0-1=1$

The orbital 2s of hydrogen like species can be diagrammatically represented as :


There is a spherical region of zero electron density between $1 s$ and the $2 s$ orbital of hydrogen like species. This spherical region is called the node. Only one radial node exist between $1 s$ and $2 s$.

The plot of wave function for $2 s$ verses radial distance from the nucleus is represented as



Point $P$ represents the region of zero electron density at a distance $r$ from the nucleus. This represents spherical node.

- Option (D) : Incorrect. The probability density at nucleus is for $2p$ orbital is zero.

The ion $\mathrm{He}^{+}$having 1 electron (in $1 s$ orbital) is hydrogen like species; hence, its $1 s$ orbital will have:

(i) Wave function same as that of hydrogen, i.e.,

$ \psi_{1,0,0}=\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} $

(ii) The probability density for the electron in the $1 s$ orbital is represented as:


For $1 s$ orbital probability density is maximum at the nucleus.

(iii) Probability density for $2 s$ orbitals is given by

$ =\frac{1}{2 a_0^{3 / 2}}\left(1-\frac{r}{2 a_0}\right) e^{-r / 2 a_0} $

Probability density $\propto \frac{1}{a_0^3}$

(iv) For electron in the $1 s$ orbital, the wave function of electron is not the function of angular wave function i.e., $\theta$ and $\phi$. Hence, 1 s orbital cannot have the wave function

$ \psi_{n, l, m_l} \propto\left(\frac{Z}{a_0}\right)^{5 / 2} r e^{\left(-Z r / 2 a_0\right) \cos \theta} $

(i) Wave function for 1s orbital of hydrogen atom with $n, l$ and $m_l$ as quantum numbers is given

$ \psi_{n, l, m_l}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} $

Where, $\mathrm{Z}$ is atomic number of the element, $r$ is the variable distance of electron from the nucleus, $a_0$ is the Bohr's radius

For $1 \mathrm{~s}$ orbital, $n=1, l=0, m_l=0$ and wave function is represented as :

$ \begin{aligned} & \psi_{1,0,0}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} \\\\ \Rightarrow \psi_{1,0,0} & \propto\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} \end{aligned} $

(ii) Energy of excitation of electron from $n_1=2$ to $n_2=4$ is given as :

$ \begin{aligned} & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\\\ & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{1}{4^2}-\frac{1}{2^2}\right) \\\\ & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{-3}{16}\right)=\frac{3 R}{16} \end{aligned} $

(iii) Energy of excitation of electron from $n_1=2$ to $n_2=4$ is given as:

$ \begin{aligned} \mathrm{E}_{2 \rightarrow 6} & \propto-\mathrm{R}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\\\ \mathrm{E}_{2 \rightarrow 6} & \propto-R\left(\frac{1}{6^2}-\frac{1}{2^2}\right) \\\\ & =-\left(\frac{1}{36}-\frac{1}{4}\right) R \\\\ E_{2 \rightarrow 6} & \propto \frac{+8 R}{36} \end{aligned} $

The ratio of the energy of transition in the two cases is given by :

$ \begin{aligned} & \frac{E_{2 \rightarrow 4}}{E_{2 \rightarrow 6}}=\frac{3 R}{16} \times \frac{6}{8 R}=\frac{27}{32} \\\\ & E_{2 \rightarrow 4}=\frac{27}{32} \times E_{2 \rightarrow 6} \end{aligned} $

the energy of transition from 2 to 4 excited state is $\frac{27}{32}$ times the energy of transition from 2 to 6 excited state.

Total number of orbitals associated with principal quantum number n is = n².

Here, n = 5.

$ \therefore $ The total number of orbitals associated with the principal quantum number 5 is = 5² = 25

We know, Kinetic Energy (K_E) = ${1 \over 2}m{v^2}$

$\therefore$ mv² = 2K_E

$ \Rightarrow $ m²v² = 2mK_E

$ \Rightarrow $ mv = $\sqrt {2m{K_E}} $

For an electron of charge 'e' which is passes through 'V' volt, kinetic energy of electron will be

K_E = eV

$\therefore$ m = $\sqrt {2meV} $

We know de-Broglie wavelength for an electron$\left( \lambda \right)$ = ${h \over p} = {h \over {mv}}$

$\therefore$ ${h \over \lambda }$ = mv = $\sqrt {2meV} $

The probability of finding an electron of hydrogen atom in a spherical shell of infinitesimal thickness, $d r$, at a distance $r$ from the nucleus, with volume $d V=4 \pi r^2 d r$ is

$ P=\psi^2 \cdot d V=\psi^2 \cdot 4 \pi r^2 d r=R^2(r) 4 \pi r^2 d r $

Here, $R^2(r)$ is the radial density function.

For $1 s$ subshell, $n=1, l=0$ and $n-l-1=0$. Thus, the number of radial and angular nodes $=0$. For 1s orbital, the probability of finding an electron will be maximum at near to the the nucleus, as 1s orbital is the nearest to the nucleus as it is the lowest orbital in terms of energy. Therefore, the plot of radial probability $P=R^2(r) 4 \pi r^2$ versus $r$ is as follows :

Energy(E_n) = $ - {{13.6{Z^2}} \over {{n^2}}}eV$

where n = 1, 2, 3, 4, ......

For hydrogen Z = 1 and excited state starts from n = 2

So putting n = 2

E₂ = $ - {{13.6} \over {{2^2}}}eV$ = -3.4 eV

The electronic configuration of Rubidium (Rb = 37):
1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹

As you can see last electron or valence electron enterin 5s subshell

So, the quantum numbers are n = 5, $l$ = 0, m = 0, s = $ \pm {1 \over 2}$

Given $E$ = $- 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)$
                = $ - 2.178 \times {10^{ - 8}}{Z^2}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$
Electron in an hydrogen atom exited from level n = 1 to n = 2. So n₂ = 2 and n₁ = 1.
And for H, Z = 1
$\therefore$ $E = - 2.178 \times {10^{ - 8}}\left( {{1 \over {{{\left( 2 \right)}^2}}} - {1 \over {{{\left( 1 \right)}^2}}}} \right)$
           = 1.6335 $ \times $ 10^-18 J

We know $E$ = ${{hc} \over \lambda }$
$\therefore$ ${{hc} \over \lambda }$ = 1.6335 $ \times $ 10^-18
then $\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}$
             $ = 1.214 \times {10^{ - 7}}m$

(a) n = 4, $l$ = 1 (p-subshell), so 4p
(b) n = 4, $l$ = 0 (s-subshell), so 4s
(c) n = 3, $l$ = 2 (d-subshell), so 3d
(d) n = 3, $l$ = 1 (p-subshell), so 3p

Accroding to the Bohr ( n + $l$ ) rule,
Enery order of the subshell : 3p < 4s < 3d < 4p

Note: When two orbital have the same value of ( n + $l$ ) then the orbital which have lower value of n will be filled first.

For Bohr orbit, angular momentum is

$mvr_n = \frac{nh}{2\pi}$

Velocity, $v = \frac{nh}{2\pi mr_n} \quad \text{... (i)}$

Kinetic energy, $K.E. = \frac{1}{2}mv^2 \quad \text{... (ii)}$

By putting the value of $v$ from (i) into (ii),

$K.E. = \frac{1}{2}m \times \frac{n^2h^2}{4\pi^2m^2r_n^2} = \frac{n^2h^2}{8\pi^2mr_n^2}$

For second Bohr orbit, $n = 2$

$r_n = a_0 \times n^2 \quad (a_0 = \text{Bohr radius})$

$r_n = 4a_0$

$K.E. = \frac{(2)^2h^2}{8\pi^2m(4a_0)^2}$

Thus, $K.E. = \frac{h^2}{32\pi^2ma_0^2}$

Here a photon is divided into two photons of different wavelength.

According to the law of conservation of energy,

Energy of photon before dividing into two parts = Energy of first photon + Energy of second photon

${{hc} \over \lambda } = {{hc} \over {{\lambda _1}}} + {{hc} \over {{\lambda _2}}}$

$ \Rightarrow {1 \over \lambda } = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$

$ \Rightarrow {1 \over {355}} = {1 \over {680}} + {1 \over {{\lambda _2}}}$

${\lambda _2} = 742.77$ nm = 743 nm

Energy required to break one Cl₂ molecule = ${{242 \times {{10}^3}} \over {6.02 \times {{10}^{23}}}}$ J

As E = ${{hc} \over \lambda }$

So $\lambda = {{hc} \over E}$

= ${{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 6.02 \times {{10}^{23}}} \over {242 \times {{10}^3}}}$

= 494 $ \times $ 10^-9 m

= 494 nm

Ionisation energy(IE) - It is the energy required to move an electron from ground state to infinity.

IE = ${E_\infty } - {E_1}$ = $0 - {E_1}$ = $ - {E_1}$

$\therefore$ E₁ of He⁺ = - 19.6 x 10^–18 J atom^–1

Energy of a species at n state,

(E_n)species = (E_n)_hydrogen $ \times $ Z²

$\therefore$ (E₁)_hydrogen = ${{ - 19.6 \times {{10}^{ - 18}}} \over 4}$ [ For He, Z = 2 ]

(E₁)_Li⁺² = ${{ - 19.6 \times {{10}^{ - 18}}} \over 4} \times {3^2}$

= -4.41 x 10^–17 J atom^–1

For hydrogen-like species, the energy levels depend on the principal quantum number $n$ and are inversely proportional to the square of $n$, and they can be characterized by their nuclear charge $Z$. The energy for a hydrogen-like ion can be expressed as:

$E_n = -\dfrac{Z^2R_H}{n^2}$

where $R_H$ is the Rydberg constant for hydrogen, $Z$ is the atomic number (for Li²⁺, $Z=3$), and $n$ is the principal quantum number associated with the energy level.

Given that S₂ has energy equal to the ground state energy of the hydrogen atom ($E_n$ for hydrogen when $n=1$ and $Z=1$ is $E_1 = -R_H$), and given that S₂ has one radial node. The radial node information tells us about the principal quantum number $n$, as the number of radial nodes is given by $n-l-1$, where $l$ is the azimuthal quantum number.

The ground state of hydrogen corresponds to $n=1$. For the given species Li²⁺ to have the same energy as the ground state of the hydrogen atom but for state S₂, we can use the energy relation. Since the energy is specified to be the same as hydrogen's ground state, let's set up the equality according to the equation given and solve for $n$ specific to the condition (Li²⁺ is considered here, but the condition is about energy equivalence).

Given that S₂ has one radial node, it cannot be the ground state energy level for Li²⁺ (which would directly correspond to $n=1$), it implies a different $n$. For one radial node, the condition $n-l-1 = 1$ must be met.

The options presented are:

• 1s ($n=1$, $l=0$) - No radial nodes
• 2s ($n=2$, $l=0$) - One radial node
• 2p ($n=2$, $l=1$) - No radial nodes, due to the different $l$
• 3s ($n=3$, $l=0$) - Two radial nodes

S₁ is described as having one radial node. Based on the rule for the number of radial nodes ($n-l-1$), the only states that fits this condition directly from the options provided are $2s$ (since for a $2s$ orbital, $n=2$, $l=0$, yielding $2-0-1=1$ radial node).

Hence, the correct option for S₁ is:

Option B: 2s

For a hydrogen-like ion, the energy levels can be given by the formula:

$E_n = -\frac{Z^2}{n^2} E_0$

where $E_n$ is the energy of the nth level, $Z$ is the atomic number (for Li$^{2+}$, $Z=3$), $n$ is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom ($-13.6 \, eV$).

Given that state $S_2$ has energy equal to the ground state energy of the hydrogen atom and one radial node, we identify that $S_2$ corresponds to $n=2$ for a hydrogen atom. This is because for hydrogen-like species, the number of radial nodes is given by $n-1$, where $n$ is the principal quantum number. The ground state ($n=1$) has 0 nodes, the first excited state ($n=2$) has 1 radial node, etc.

Since the energy of $S_2$ is equal to the ground state energy of the hydrogen atom, we can directly compare the energies. For hydrogen ($Z=1$), the ground state energy ($n=1$) is:

$E = -E_0$

For the Li$^{2+}$ ion in state $S_2$ (which we established is equivalent to $n=2$ in terms of energy for hydrogen), we use the formula $E_n = -\frac{Z^2}{n^2} E_0$. Since $Z=3$ for Li$^{2+}$, and given that $S_2$ has the energy equivalent to the ground state of hydrogen ($E_0$), we solve for the energy ratio rather than the specific energy of $S_2$.

We then look at $S_1$, which we know must be the ground state for Li$^{2+}$ since it is the state before $S_2$ and has one radial node (indicating $n=2$ for $S_1$).

Thus, for $S_1$, which actually corresponds to $n=2$ for Li$^{2+}$, the energy in units of the hydrogen atom ground state energy is:

$E_{S_1} = -\frac{Z^2}{n^2} E_0 = -\frac{3^2}{2^2}E_0 = -\frac{9}{4}E_0$

Now, to express this in units of the hydrogen atom ground state energy ($-E_0$):

$\frac{E_{S_1}}{E_0} = -\frac{9}{4} = -2.25$

So, considering the provided options and the fact that energy levels are usually considered in positive values when comparing magnitudes, the correct answer is:

Option C: 2.25.

To identify the orbital angular momentum quantum number, $l$, for the state $S_2$ of a hydrogen-like species such as $Li^{2+}$, we can refer to the given information and the known equations for the energy levels of hydrogen-like atoms. Hydrogen-like atoms or ions have only one electron and their energy in a particular state is given by the equation:

$E_n = -\frac{Z^2}{n^2} E_0$

where:

• $E_n$ is the energy of the electron in the nth energy level,
• $Z$ is the atomic number of the species (for $Li$, $Z = 3$),
• $n$ is the principal quantum number,
• $E_0$ is the energy of the ground state of hydrogen ($-13.6$ eV).

The problem states that $S_2$ has energy equal to the ground state energy of the hydrogen atom. The ground state of hydrogen corresponds to $n = 1$ and $E_0 = -13.6$ eV. However, for $Li^{2+}$, with $Z = 3$, the same energy level could be attained at a different value of $n$ since the $Z^2$ factor magnifies the energy levels with increasing atomic number. Let's calculate the principal quantum number, $n$, for the $Li^{2+}$ ion that would give it an energy equal to $E_0$:

For $Li^{2+}$ ion to have the ground state energy of a hydrogen atom, we can set the energies equal and solve for $n$:

$-\frac{Z^2}{n^2}E_0 = E_0$

Substituting $Z = 3$ and simplifying:

$-\frac{9}{n^2} = 1$

From which we find, $n^2 = 9$ and thus $n = 3$.

Furthermore, the problem states that $S_1$ has one radial node and upon absorbing light, it transitions to $S_2$ which also has one radial node. Radial nodes are related to the principal quantum number $n$ and the angular quantum number $l$ by the formula:

Number of radial nodes = $n - l - 1$

Given that $S_2$ has one radial node, we can plug $n = 3$ into the radial node formula to solve for $l$:

$1 = 3 - l - 1$

This simplifies to:

$l = 3 - 2$

$l = 1$

Therefore, the orbital angular momentum quantum number of the state $S_2$ is 1, which corresponds to option B.

% error in velocity = ${{\Delta V} \over V} \times 100$

$ \therefore $ 0.005 = ${{\Delta V} \over {600}} \times 100$

$ \Rightarrow $ $\Delta $V = 3 $ \times $ 10^-2

According to Heisenberg uncertainty principle,

$\Delta x.m\Delta V \ge {h \over {4\pi }}$

$ \Rightarrow $ $\Delta x = {h \over {4\pi m\Delta V}}$

$ \Rightarrow $ $\Delta x = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^{ - 2}}}}$

             = 1.92 $ \times $ 10^-3 m

Wavelength$\left( \lambda \right)$ = ${h \over {mv}}$

= ${{6.63 \times {{10}^{ - 34}}} \over {1.67 \times {{10}^{ - 27}} \times {{10}^3}}}$

= 0.4 $\times$ 10^-9

= 0.4 nm

Species	No. of electrons
C22-	12+2 = 14
O2-	16+1 = 17
CO	6+8 = 14
NO	7+8 = 15
NO+	7+8-1 = 14
C22-	12+2 = 14
CN-	6+7+1 = 14
N2	14
CN-	6+7+1 = 14
N2	14
O22-	16+2 = 18
C2	12
N2	14
O2-	16+1 = 17
NO+	7+8-1 = 14
CO	6+8 = 14

So, option $(B)$ is correct.

Note :

1 eV/atom = 96.485 $ \times $ 10³ J/mol

$\therefore$ 13.6 eV/atom = 13.6$ \times $ 96.485 $ \times $ 10³ J/mol = 1.312 × 10⁶ J mol⁻¹

Energy required to excite the electron from n₁ to n₂ is

$\Delta E = 13.6 \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

= 1.312 × 10⁶ × 1$\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$

= 1.312 × ${3 \over 4}$ × 10⁶

= 9.84 × 10⁵ J mol⁻¹

(A) Orbital angular momentum of the electron in a hydrogen-like atomic orbital is represented by azimuthal quantum number.

(B) A hydrogen-like one-electron wave function obeying Pauli principle correspond to Electron spin quantum number.

(C) Shape, size and orientation of hydrogen-like atomic orbitals are determined by Azimuthal, principal and Magnetic quantum number.

(D) Probability density of electron at the nucleus in hydrogen-like atom is determined by Principal quantum number, Azimuthal quantum number and Magnetic quantum number.

With increase in the atomic number, the proton-proton electrostatic repulsion begins to overcome attractive forces involving proton and neutrons in heavier nuclides. The stability relationship can be represented by a line with a slope of 45$^\circ$, i.e., the maximum stability is attained when N = Z. Right of the curve a radioactive nuclide would be neutron rich and would decay by $\beta$-emission to produce a daughter nucleus with a lower n/p ratio. For heavier nuclides, p-p repulsions start to offset the attractive forces and an excess of neutrons over protons, is required for stability.

In which quantum number have highest ( n + l ) value, will represent highest energy of an atom.

In option (C), n + l = 3 + 2 = 5 = maximum.

Given:

A positron is emitted from:

$ {}^{23}_{11}\text{Na} $

Step 1: Recall what happens in positron emission.

A positron is a $\beta^+$ particle, i.e. $ {}^0_{+1}e $.

In positron emission:

$ p \rightarrow n + e^+ + \nu_e $

So inside the nucleus:

• Number of protons decreases by 1,

• Number of neutrons increases by 1,

• The mass number (A) remains the same.

Step 2: Apply this to sodium-23.

$ {}^{23}_{11}\text{Na} \rightarrow {}^{23}_{10}\text{X} + e^+ + \nu_e $

So the new nucleus has:

• Atomic number $Z = 10$

• Mass number $A = 23$

The element with $Z = 10$ is Neon (Ne).

Hence, the daughter nucleus is $ {}^{23}_{10}\text{Ne} $.

Step 3: Find the required ratio.

They ask for the ratio of atomic mass (mass number) to atomic number of the resulting nuclide:

$ \frac{A}{Z} = \frac{23}{10} $

✅ Final Answer:

$ \boxed{\frac{23}{10}} $

Option C: $\dfrac{23}{10}$

Option $(c)$ is correct.

In $\,\,{K^ + }\left( {19} \right)\,\,$ no. of electrons

$ = 19 - 1\,\,\,\,\, = 18$

In $\,\,{Cl^ - }\left( {17} \right)\,\,$ no. of electrons

$ = 17 + 1\,\,\,\,\, = 18$

In $\,\,{Ca^ 2+ }\left( {20} \right)\,\,$ no. of electrons

$ = 20 - 2\,\,\,\,\, = 18$

In $\,\,{Sc^ 3+ }\left( {21} \right)\,\,$ no. of electrons

$ = 21 - 3\,\,\,\,\, = 18$

Formula of angular momentum of an electron,

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

here n = 5

$\therefore$ mvr = ${{5h} \over {2\pi }}$ = 2.5${{h} \over {\pi }}$

% error in velocity = ${{\Delta V} \over V} \times 100$

$ \therefore $ 0.001 = ${{\Delta V} \over {300}} \times 100$

$ \Rightarrow $ $\Delta $V = 3 $ \times $ 10^-3

According to Heisenberg uncertainty principle,

$\Delta x.m\Delta V \ge {h \over {4\pi }}$

$ \Rightarrow $ $\Delta x = {h \over {4\pi m\Delta V}}$

$ \Rightarrow $ $\Delta x = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^{ - 3}}}}$

             = 1.92 $ \times $ 10^-2 m

(A) The $\mathrm{V}_n$ and $\mathrm{K}_n$ denotes potential energy and kinetic energy of an electron.

The expression for kinetic theory and potential energy are given by Bohr atomic theory as shown below.

$ \begin{array}{rlrl} & \mathrm{V}_n =-\frac{\mathrm{Ze}^2}{r} \text { and } \mathrm{K}_n=\frac{\mathrm{Ze}^2}{2 r} \\ \text { Hence, } & \frac{\mathrm{V}_n}{\mathrm{~K}_n} =\frac{-\frac{\mathrm{Ze}^2}{r}}{\frac{\mathrm{Ze}^2}{2 r}} \\ & \therefore \frac{\mathrm{~V}_n}{\mathrm{~K}_n} =-2 \end{array} $

Hence, option A in column I matches with option R in column II.

(B) The $\mathrm{E}_n$ denoted the total energy of an electron which is the sum of potential and kinetic energy of an electron.

The radius of $n^{\text {th }}$ orbit of an atom is denoted by $r_n$.

Where $n$ denoted the number of an orbit.

The radius of $n^{\text {th }}$ orbit is directly proportional to the square of $n$ and the total energy of an electron is inversely proportional to that of square of $n$.

$ r_n \propto n^2 \text { and } \mathrm{E}_n \propto \frac{1}{n^2} $

Therefore,

$ \mathrm{E}_n \propto \frac{1}{r_n} \text { or } r_n \propto \mathrm{E}_n^{-1} $

The value of $x$ will be,

$ x=-1 $

Hence, option B in column I matches with option Q in column II.

(C) According to the Bohr theory, angular momentum for an orbital is given as,

$ \omega=\sqrt{l(l+1)} \frac{h}{2 \pi} $

Here, $\omega$ is the angular momentum for an orbital, $l$ is the azimuthal quantum number and $h$ is the Planck's constant.

The lowest atomic orbital in any atom is $1 s$, which has least energy in the atom.

For $1 s$ orbital, the azimuthal quantum number is 0 as the orbital possess spherical shape.

Substituting the value of $l$ in above equation for 1s orbital,

$ \begin{aligned} & \omega=\sqrt{0(0+1)} \frac{h}{2 \pi} \\ & \omega=0 \end{aligned} $

Hence, the value of angular momentum for lowest orbital will be zero.

Hence, option C in column I matches with option P in column II.

(D) The Z denotes the atomic number of an element.

The radius of an orbit and atomic number of an element are related by following relation.

$ \frac{1}{r_n} \propto Z_n $

If $\frac{1}{r_n} \propto Z^y$, then $y=1$

Hence, option D in column I matches with option $S$ in column II.

As here is no electric and magnetic field so ignore m to calculate the energy of orbital.

As here atom is multi-electron so ( n + l ) rule is applicable here. This rule says those orbitals which have more value of ( n + l ) will have more energy.

In (A), n + l = 1 + 0 = 1

In (B), n + l = 2 + 0 = 2

In (C), n + l = 2 + 1 = 3

In (D), n + l = 3 + 2 = 5

In (E), n + l = 3 + 2 = 5

So (D) and (E) will be of equal energy.

Ions	No. of electrons
CH3+	8
H3O+	10
NH3	10
CH3-	10

$\therefore$ $\,\,\,$ ${H_3}{O^ + },N{H_3}$ and $C{H_3}^ - $ are isoelectronic.

In $SO_3^{2 - }$ no of electrons = 16 + 24 + 2 = 42

In $CO_3^{2 - }$ no of electrons = 6 + 24 + 2 = 32

In $NO_3^{-}$ no of electrons = 7 + 24 + 1 = 32

So they are not isoelectronic.

To find the number of radial nodes in an orbital, we use the formula:

$\text{Number of radial nodes} = n - l - 1$

where $n$ is the principal quantum number, and $l$ is the azimuthal quantum number (also known as the angular momentum quantum number).

For the 3s orbital:

$n = 3$ (since it is the 3rd energy level) and $l = 0$ (s orbital corresponds to $l = 0$).

Plugging these values into the formula gives:

$\text{Number of radial nodes for 3s} = 3 - 0 - 1 = 2$

For the 2p orbital:

$n = 2$ (since it is the 2nd energy level) and $l = 1$ (p orbital corresponds to $l = 1$).

Therefore, the number of radial nodes for the 2p orbital is:

$\text{Number of radial nodes for 2p} = 2 - 1 - 1 = 0$

Thus, the number of radial nodes of 3s and 2p orbitals are respectively 2 and 0.

Option A is correct: 2, 0

For 4f orbital,

n = 4 and l = 3

Note : ( for s orbital l = 0, for p orbital l = 1, for d orbital l = 2, for f orbital l = 3)

Value of m = + l to - l

Here m = +3 to -3

Value of s = $ + {1 \over 2}$ or $ - {1 \over 2}$

Electronic configuration of Cr (Z = 24) =

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵

For 2p⁶ and 3p⁶ , l = 1. Here in 2p and 3p orbital total 6 + 6 = 12 electrons present.

For 3d⁵ , l = 2. Here in 3d orbital total 5 electrons present.

We know Rydberg formula,

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

[ for hydrogen atom Z = 1 ]

     = $1.097 \times {10^7}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$

$ \therefore $ $\lambda $ = ${1 \over {1.097 \times {{10}^7}}}$

= 9.11 $ \times $10^-8 m

= 91.1 $ \times $10^-9 m

= 91.1 nm

[ as 1 nm = 10^-9 m ]

Isoelectronic means those species whose no of electrons are same.

Ions	Atomic No	No of Electron
K+	19	19-1=18
Ca+2	20	20-2=18
Sc+3	21	21-3=18
Cl-	17	17+1=18

So, option $(C)$ is correct.

To compare the radius of these different orbits to the first Bohr's orbit of the hydrogen atom, we can use the formula for the radius of the nth orbit in the hydrogen-like atom, which is given by:

$ r_n = \frac{n^2h^2}{4\pi^2kme^2}Z^{-1} $

Where:

• $r_n$ is the radius of the nth orbit,
• $n$ is the principal quantum number,
• $h$ is Planck's constant,
• $k$ is Coulomb's constant,
• $m$ is the mass of the electron,
• $e$ is the charge of the electron,
• $Z$ is the atomic number or the number of protons in the nucleus (which also equals the charge of the nucleus in a hydrogen-like atom).

For the first Bohr orbit of the hydrogen atom ($n=1$ and $Z=1$), the radius is:

$ r_1 = \frac{1^2h^2}{4\pi^2kme^2}(1)^{-1} $

We must compare this with the options provided. The formula simplifies to identifying where the $n^2/Z$ ratio for each option is equal to 1, as it is for hydrogen's first orbit.

• Option A: He⁺ ($Z=2$, $n=2$)

  $\frac{n^2}{Z} = \frac{2^2}{2} = 2$

• Option B: Li²⁺ ($Z=3$, $n=2$)

  $\frac{n^2}{Z} = \frac{2^2}{3} \approx 1.33$

• Option C: Li²⁺ ($Z=3$, $n=3$)

  $\frac{n^2}{Z} = \frac{3^2}{3} = 3$

• Option D: Be³⁺ ($Z=4$, $n=2$)

  $\frac{n^2}{Z} = \frac{2^2}{4} = 1$

Thus, the radius is the same as that of the first Bohr's orbit of hydrogen atom only for Option D: Be³⁺ ($n=2$), because in this case, the ratio $n^2/Z = 1$, matching the ratio for hydrogen's first orbit.

Isoelectronic means those species whose electron number are same.

Species	Atomic Number	Electron Number
N3-	7	7 + 3 = 10
F-	9	9 + 1 = 10
Na+	11	11 - 1 = 10
Be	4	4
Al3+	13	13 - 3 = 10
Cl-	17	17 + 1 = 8
Ca2+	20	20 - 2 = 18
Cs+	55	55 - 1 = 54
Br	35	35
Mg2+	12	12 -2 = 10

$\therefore$ N^3-, F^-, Na⁺

As per de-Broglie wavelength of a particle is

Wavelength ($\lambda $) = ${h \over {mv}}$

Where h is plank's constant, m is mass of the particle and v is the velocity of the particle

$\therefore \lambda $ = ${{6.63 \times {{10}^{ - 34}}} \over {0.06 \times 10}}$

= ${{6.63 \times {{10}^{ - 34}} \times 10} \over {0.6 \times 10}}$

= ${{6.63 \times {{10}^{ - 33}}} \over 6}$

$ \cong 10{}^{ - 33}$