Structure of Atom

$ \text { (1) Wavelength of } \mathrm{A}=400 \mathrm{~nm} \text {. } $

$ \text { (2) For photon } \mathrm{B}, \ \lambda=\frac{\mathrm{C}}{v}=\frac{3 \times 10^8}{10^{16}} $

$ =3 \times 10^{-8}\ \text{m} =30 \times 10^{-9}\ \text{m} =30 \mathrm{~nm} . $

(3) For photon $\mathrm{C}$, wave number is given, and $\bar{v}=\frac{1}{\lambda}$. So $ \mathrm{C}(\lambda)=\frac{1}{\bar{v}}=\frac{1}{10^4}=10^{-4} \mathrm{~cm} =10^{-6} \mathrm{~m}=1000 \mathrm{~nm}$

Now compare the wavelengths: $\lambda_{\mathrm{C}}>\lambda_{\mathrm{A}}>\lambda_{\mathrm{B}}$

Energy of a photon is inversely proportional to wavelength: Energy(E) $ \propto \frac{1}{\lambda}$

So, smaller wavelength means higher energy. Therefore: $\mathrm{E}_{\mathrm{B}}>\mathrm{E}_{\mathrm{A}}>\mathrm{E}_{\mathrm{C}}$

Balmer series corresponds to transitions ending at $n_1=2$.

For hydrogen, the wave number is given by

$ \bar{\nu}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \quad n_2>n_1 $

For Balmer series: $n_1=2$, so

$ \bar{\nu}=R\left(\frac{1}{4}-\frac{1}{n_2^2}\right), \quad n_2=3,4,5,\dots $

Now compute the first few Balmer lines:

1. For $n_2=3$:

$ \bar{\nu}=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{9-4}{36}\right)=\frac{5R}{36} $

2. For $n_2=4$:

$ \bar{\nu}=R\left(\frac{1}{4}-\frac{1}{16}\right)=R\left(\frac{4-1}{16}\right)=\frac{3R}{16} $

3. For $n_2=5$:

$ \bar{\nu}=R\left(\frac{1}{4}-\frac{1}{25}\right)=R\left(\frac{25-4}{100}\right)=\frac{21R}{100} $

These three values $\left(\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100}\right)$ match Option A.

Correct option: A

The nodal surface (as seen in Figure 1) is the place where the probability density becomes zero.

For an $s$-orbital, this nodal surface is a spherical node (a spherical shell around the nucleus).

At a spherical (radial) node, the radial wave function becomes zero.

$ \psi_{\mathrm{r}}=0 $

So, in Figure 2, the correct point is where the 2s wave function curve cuts the horizontal axis (changes sign). That point represents the spherical node shown in Figure 1.

$\mathrm{n}_1 \rightarrow$ lower energy level

$\mathrm{n}_2 \rightarrow$ higher energy level

$ \begin{aligned} & \mathrm{n}_1+\mathrm{n}_2=4, \quad \mathrm{n}_2=3 \\ & \mathrm{n}_2-\mathrm{n}_1=2, \quad \mathrm{n}_1=1 \end{aligned} $

Rydberg's formula :

$ \begin{aligned} & \frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \mathrm{Z}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}(3)^2\left[\frac{1}{1^2}-\frac{1}{3^2}\right] \\ & \frac{1}{\lambda}=8 \mathrm{R}_{\mathrm{H}} \\ & \lambda=\frac{1}{8 \mathrm{R}_{\mathrm{H}}} \\ & \lambda=\frac{1}{8 \times 1.1 \times 10^5} \\ & \lambda=\frac{1000}{8.8} \times 10^{-8} \mathrm{~cm} \\ & \lambda=113.63 \times 10^{-8} \mathrm{~cm} \\ & \lambda \simeq 1.14 \times 10^{-6} \mathrm{~cm} \end{aligned} $

Let the work functions be $\phi_A$ and $\phi_B$.

Given ratio of work functions:

$\phi_A : \phi_B = 1:2 \;\Rightarrow\; \phi_B = 2\phi_A$

Photon energy is $h\nu = 6\,\text{eV}$.

By Einstein’s photoelectric equation:

$K_{\max} = h\nu - \phi$

So,

$K_A = 6 - \phi_A$

$K_B = 6 - \phi_B = 6 - 2\phi_A$

Given,

$\frac{K_A}{K_B} = \frac{2.642}{1} = 2.642$

So,

$\frac{6-\phi_A}{6-2\phi_A} = 2.642$

Cross-multiplying:

$6 - \phi_A = 2.642(6 - 2\phi_A)$

$6 - \phi_A = 15.852 - 5.284\phi_A$

Bring like terms together:

$5.284\phi_A - \phi_A = 15.852 - 6$

$4.284\phi_A = 9.852$

$\phi_A = \frac{9.852}{4.284} \approx 2.3\,\text{eV}$

Then,

$\phi_B = 2\phi_A \approx 4.6\,\text{eV}$

Answer: $\boxed{2.3\,\text{eV},\; 4.6\,\text{eV}}$ (Option C)

Statement A

For ${}_{12}^{24}\mathrm{Mg}$:

• Atomic number $Z=12$ $\Rightarrow$ number of protons $=12$

• Mass number $A=24$ $\Rightarrow$ protons + neutrons $=24$

So, neutrons $=24-12=12$.

Given statement says 24 protons and 12 neutrons, which is wrong.

✅ A is incorrect

Statement B

Wavelength $\lambda$ is given by

$ \lambda=\frac{c}{\nu} $

Here $c=3\times 10^8\ \text{m s}^{-1}$ and $\nu=4.5\times 10^{15}\ \text{s}^{-1}$

$ \lambda=\frac{3\times 10^8}{4.5\times 10^{15}} =0.6667\times 10^{-7} =6.67\times 10^{-8}\ \text{m} $

Matches the given value.

✅ B is correct

Statement C

Energy of radiation:

$ E=\frac{hc}{\lambda}\Rightarrow E\propto \frac{1}{\lambda} $

So,

$ \frac{E_1}{E_2}=\frac{\lambda_2}{\lambda_1}=\frac{300}{900}=\frac{1}{3} $

Therefore,

$ E_1:E_2=1:3 $

But given is $E_1:E_2=3:1$, which is wrong.

✅ C is incorrect

Statement D

Energy per photon:

$ E=\frac{hc}{\lambda} $

Given $\lambda=2000\ \text{pm}=2000\times 10^{-12}=2\times 10^{-9}\ \text{m}$

$ E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{2\times 10^{-9}} =9.939\times 10^{-17}\ \text{J} $

Number of photons for $1\ \text{J}$:

$ N=\frac{1}{9.939\times 10^{-17}}=1.006\times 10^{16} $

Matches the given value.

✅ D is correct

Incorrect statements: A and C only

Correct option: Option B

(A) $n=5$. So we look at all subshells for this value of $n$, i.e. $l=0,1,2,3,4$, and check where the given magnetic quantum number $m_l=-1$ is possible.

For $l=0 {~m}_l=0$, i.e. only one orbital is possible and it has $m_l=0$, so here we do not get any orbital with $m_l=-1$.

For $l=1 {~m}_l=-{1}, 0,1 \Rightarrow 2$ electrons can have $m_l=-1$ in this subshell, because there is exactly one orbital with $m_l=-1$, and one orbital can hold 2 electrons (with opposite spins).

For $l=2 {~m}_l=-2,-{1}, 0,1,2 \Rightarrow 2$ electrons can again have $m_l=-1$, because here also there is one orbital with $m_l=-1$, which can contain 2 electrons.

For $l=3 {~m}_l=-3,-2,-{1}, 0,1,2,3 \Rightarrow 2$ electrons can have $m_l=-1$ for the same reason: one orbital with this $m_l$ value, accommodating 2 electrons.

For $l=4 {~m}_l=-4,-3,-2,{1}, 0,1,2,3,4 \Rightarrow 2$ electrons can have the given magnetic quantum number in this subshell as well (one such orbital $\Rightarrow$ 2 electrons).

Total number of electrons $=8$, because there are 4 such orbitals (for $l=1,2,3,4$), each holding 2 electrons.

(B) ${n}=3, {l}=2, {~m}_l=-1, {~m}_{{s}}=+1 / 2 \Rightarrow$ only 1 electron is possible, because all four quantum numbers are completely specified. By Pauli’s exclusion principle, no two electrons in an atom can have the same set of four quantum numbers, so at most one electron can have this exact combination.

For one–electron (hydrogen-like) systems, the energy of the $n^{\text{th}}$ stationary state is (NCERT):

$ E_n = -2.18 \times 10^{-18}\,\frac{Z^2}{n^2}\ \text{J} $

where $Z$ is the atomic number and $n$ is the orbit number.

Check each option

Option A: Second orbit of $\mathrm{He}^+$

Here $Z=2,\ n=2$

$ E_2 = -2.18\times 10^{-18}\frac{2^2}{2^2} = -2.18\times 10^{-18}\ \text{J} $

Given is $+2.18\times 10^{-18}$ J (positive), so false.

Option B: First orbit of $\mathrm{He}^+$

$Z=2,\ n=1$

$ E_1 = -2.18\times 10^{-18}\frac{4}{1} = -8.72\times 10^{-18}\ \text{J} $

Given is $+8.72\times 10^{-18}$ J, so false.

Option C: Second orbit of H atom

$Z=1,\ n=2$

$ E_2 = -2.18\times 10^{-18}\frac{1}{4} = -0.545\times 10^{-18}\ \text{J} $

Given is $-1.09\times 10^{-18}$ J, so false.

Option D: Third orbit of $\mathrm{Li}^{2+}$

$Z=3,\ n=3$

$ E_3 = -2.18\times 10^{-18}\frac{9}{9} = -2.18\times 10^{-18}\ \text{J} $

This matches the given value, so true.

Correct option: D

Balmer series means transitions ending at $n=2$.

Photon energy emitted in a transition $n \to 2$ is

$ E = 13.6\,\text{eV}\left(\frac{1}{2^2}-\frac{1}{n^2}\right) $

First (lowest) Balmer line

This is the first line of Balmer series: $3 \to 2$

$ E_1 = 13.6\left(\frac{1}{4}-\frac{1}{9}\right) =13.6\left(\frac{5}{36}\right) $

Given $E_1 = x$ J.

Second Balmer line

This is $4 \to 2$

$ E_2 = 13.6\left(\frac{1}{4}-\frac{1}{16}\right) =13.6\left(\frac{3}{16}\right) $

Ratio

$ \frac{E_2}{E_1} = \frac{\left(\frac{3}{16}\right)}{\left(\frac{5}{36}\right)} = \frac{3}{16}\cdot\frac{36}{5} = \frac{108}{80} = 1.35 $

So,

$ E_2 = 1.35\,x $

Correct option: B) $1.35x$

$\begin{aligned} & E_n=-R_H \times \frac{Z^2}{n^2} \\ & \Delta E=2.18 \times 10^{-11} \times 10^{-7} \times 1^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\ & =1.635 \times 10^{-18} \mathrm{Joule} / \text { atom } \\ & =1.635 \times 10^{-18} \times 6.02 \times 10^{23} \text { Joule } / \text { mole } \\ & =9.835 \times 10^5 \text { Joule } / \text { mole }\end{aligned}$

For hydrogen spectral lines (emission),

$E=\frac{hc}{\lambda} \;\;\Rightarrow\;\; E \propto \frac{1}{\lambda}$

Using Rydberg formula (NCERT):

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

So photon energy is proportional to:

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

Now identify each line:

• Paschen series: $n_1=3$

• A (first line): $n_2=4 \to 3$

• C (third line): $n_2=6 \to 3$

• Balmer series: $n_1=2$

• B (second line): $n_2=4 \to 2$

• Brackett series: $n_1=4$

• D (fourth line): $n_2=8 \to 4$

Compute $\Delta$ for each:

D: $8 \to 4$

$\Delta_D=\frac{1}{4^2}-\frac{1}{8^2}=\frac{1}{16}-\frac{1}{64}=\frac{3}{64}=0.046875$

A: $4 \to 3$

$\Delta_A=\frac{1}{3^2}-\frac{1}{4^2}=\frac{1}{9}-\frac{1}{16}=\frac{7}{144}\approx 0.04861$

C: $6 \to 3$

$\Delta_C=\frac{1}{3^2}-\frac{1}{6^2}=\frac{1}{9}-\frac{1}{36}=\frac{1}{12}\approx 0.08333$

B: $4 \to 2$

$\Delta_B=\frac{1}{2^2}-\frac{1}{4^2}=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}=0.1875$

Ascending order of energy (smallest $\Delta$ to largest $\Delta$):

$D < A < C < B$

Correct option: C

Statement I: True.

When electric discharge is passed through gaseous hydrogen, $H_2$ molecules dissociate into hydrogen atoms. These atoms get excited and, on returning to lower energy levels, emit radiation of discrete frequencies, giving a line spectrum (as per NCERT description of atomic spectra).

Statement II: True.

Use the Rydberg formula (NCERT) for hydrogen-like species:

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

(a) Second line of Balmer series of $\mathrm{He}^+$

Balmer series means $n_1=2$.

Second line means transition $n_2=4 \to n_1=2$.

For $\mathrm{He}^+$, $Z=2$.

$\tilde{\nu}_{\mathrm{He}^+}=R(2)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right) =4R\left(\frac{1}{4}-\frac{1}{16}\right) =4R\left(\frac{3}{16}\right) =\frac{3R}{4}$

(b) First line of Lyman series of hydrogen atom

Lyman series means $n_1=1$.

First line means transition $n_2=2 \to n_1=1$.

For hydrogen, $Z=1$.

$\tilde{\nu}_{H}=R(1)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right) =R\left(1-\frac{1}{4}\right) =\frac{3R}{4}$

So, $\tilde{\nu}_{\mathrm{He}^+}=\tilde{\nu}_{H}$, hence their frequencies are equal.

Correct option: A

Both Statement I and Statement II are true.

$\begin{aligned} &\text { Element with atomic number } 9 \text { is Fluorine }\\ &F(9)=1 s^2 2 s^2 2 p^5 \end{aligned}$

(A) 5 electrons can be up-spin $\left[\mathrm{m}_{\mathrm{s}}=+\frac{1}{2}\right]$ and 4 electrons can be down spin $\left[\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}\right]$

(B) Unpaired electron can be in anyone of $p_x, p_y$ or $p_z$ orbital

(C) Last electron is in 2 p subshell with $\mathrm{n}=2, \ell=1$

(D) Angular node for s-orbital $=0$ while of each p -orbital $=1$

Sum of all angular node $=3$

The enhanced stability of a half-filled subshell is attributed to the following factors:

Symmetrical Distribution of Electrons: Electrons are evenly distributed across the available orbitals, leading to a balanced and stable arrangement.

Large Exchange Energy: The energy due to the exchange of electrons with parallel spins among degenerate orbitals is maximized, contributing to increased stability.

Smaller Coulombic Repulsion: The repulsion between electrons is minimized in a half-filled configuration, reducing the overall energy of the subshell.

Smaller Shielding of Electrons: In a half-filled subshell, electrons are more exposed to the nuclear charge, as there is less shielding effect from other electrons, enhancing stability.

To determine the correct statements regarding the threshold frequency of cesium, first note that the threshold frequency given is $5.16 \times 10^{14} \, \text{Hz}$. Using this frequency, we can calculate the corresponding wavelength, $\lambda$, using the equation:

$ \lambda = \frac{c}{v} = \frac{3 \times 10^8 \, \text{m/s}}{5.16 \times 10^{14} \, \text{Hz}} $

This calculation gives $\lambda \approx 581.39 \, \text{nm}$.

Key points:

The wavelength calculated is near the yellow light region. Therefore, yellow light has enough energy to cause a photoelectric effect.

When the intensity of light (brightness) decreases, the photocurrent will also decrease, as fewer photons are available to eject electrons.

Red light, with its larger wavelength and lower frequency, does not have enough energy to cause the photoelectric effect in cesium.

Blue light has a higher frequency than yellow light ($v_{\text{blue}} > v_{\text{yellow}}$), meaning it can produce a photoelectric current.

White light includes all visible frequencies, including those above the threshold, so it will also emit a current.

Thus, the correct statements involve using yellow, blue, and white light, without producing an effect with red light:

A: Yellow light produces a current.

B: Dimming yellow light reduces the current.

D: Blue light produces a current.

E: White light produces a current.

Therefore, the correct statements are A, B, D, and E.

The ground state electron configuration of chromium (Cr, $ Z = 24 $) is:

$ \text{Cr: } [\text{Ar}]\,3d^5\,4s^1 $

Let's break down the configuration:

The noble gas core $[\text{Ar}]$ represents:

$1s^2$ (with $l=0$)

$2s^2$ (with $l=0$)

$2p^6$ (with $l=1$)

$3s^2$ (with $l=0$)

$3p^6$ (with $l=1$)

Electrons with azimuthal quantum number $l = 1$ are found in $p$ orbitals:

$2p^6$: 6 electrons

$3p^6$: 6 electrons

Total $p$-electrons: $6 + 6 = 12$.

Electrons with azimuthal quantum number $l = 2$ are found in $d$ orbitals:

$3d^5$: 5 electrons

Thus, there are 12 electrons with $l = 1$ and 5 electrons with $l = 2$.

The correct option is Option B.

1. $\Psi^2=$ Probability density is maximum at nucleus.

2. Electron can exist upto infinity from nucleus.

3. True

4. Energy of electron is maximum at infinite distance from nucleus.

The orbital angular momentum for electrons in orbitals is given by the formula:

$ \text{Orbital angular momentum} = \sqrt{\ell(\ell+1)} \frac{\mathrm{h}}{2 \pi} $

For the 2s orbital, the quantum number $\ell$ is 0:

Orbital angular momentum = $\sqrt{0(0+1)} \frac{\mathrm{h}}{2 \pi} = 0$

For the 2p orbital, the quantum number $\ell$ is 1:

Orbital angular momentum = $\sqrt{1(1+1)} \frac{\mathrm{h}}{2 \pi} = \sqrt{2} \frac{\mathrm{h}}{2 \pi}$

The postulate that contradicts quantum mechanics is

Option A

“The electron in a H atom’s stationary state moves in a circle around the nucleus.”

In quantum mechanics an electron in a stationary state is described by a delocalized wave‐function (an orbital), not by a definite circular orbit.

(A) Azimuthal quantum number ( $\ell$ ) indicates the shape of orbital occupied by the electron

(C) The + and - sign in the wave function of $2 p_x$ orbital refer to the sign (Phase) of the wave function, not the charge

(D) The wave function of $2 p_x$ orbital will be zero in yz plane (Nodal plane).

In Bohr’s model the radius of the $n$th orbit is

$r_n = n^2\,a_0$

where $a_0$ is the Bohr radius. Hence for any two orbits $m$ and $n$:

$\frac{r_m}{r_n} = \frac{m^2}{n^2}\,. $

Let’s check each option:

A: $\displaystyle \frac{r_4}{r_2} = \frac{4^2}{2^2} = \frac{16}{4} = 4$ → correct

B: $\displaystyle \frac{r_8}{r_4} = \frac{8^2}{4^2} = \frac{64}{16} = 4$ → correct

C: $\displaystyle \frac{r_6}{r_4} = \frac{6^2}{4^2} = \frac{36}{16} = 2.25\;\bigl(\neq3\bigr)$ → incorrect

D: $\displaystyle \frac{r_3}{r_1} = \frac{3^2}{1^2} = 9$ → correct

Therefore the incorrect statement is Option C.

$E\,vs\,Z\,plot\,(cons\tan t\,n)\left\{ \matrix{ E - Energy\,of\,the\,stationary\,state \hfill \cr Z - atomic\,number \hfill \cr n - \Pr incipal\,quantum\,number \hfill \cr} \right.$

The formula of energy of hydrogen like species is

$E = - 13.6{{{z^2}} \over {{n^2}}}eV$

$E \propto - {{{z^2}} \over {{n^2}}}$

For constant n, $E \propto - {Z^2}$

Here, E and Z has a quadratic relationship. Since the equation (proportionality) includes a negative sign, the graph of E vs Z will be a downward - opening parabola.

The graph is like an upside down U. In the given options, graph 4 is like half upside down parabola.

The shape is characteristic of negative quadratic function, where the curve opens downward, with the vertex of the parabola at the origin.

So, the answer is option (4).

Statement I : Correct

This statement is describing the uncertainty principle.

It states that it is impossible to measure both the position and the momentum of an object.

Statement II : Correct.

If the uncertainty in the measurement of position ($\Delta x$) and the uncertainty in the measurement of momentum ($\Delta p$) are equal for an electron, then the uncertainty in the measurement of velocity ($\Delta v$) is greater than or equal to $\sqrt{\frac{h}{\pi}}\frac{1}{2m}$

This is derived from the Heisenberg uncertainty principle, $\Delta x\,.\,\Delta {p_x} \ge {h \over {4\pi }}$

Given, $\Delta x = \Delta p$

Substitute this in $\Delta x\,.\,\Delta p \ge {h \over {4\pi }}$ as

$\Delta p\,.\,\Delta p \ge {h \over {4\pi }}$

$\Delta {p^2} \ge {h \over {4\pi }}$

$\Delta {p^2} \ge {h \over {4\pi }}$

$\left\{ \matrix{ The\,formula\,for\,momentum\,(P)\,is\,P = mV \hfill \cr For\,uncerta{\mathop{\rm int}} y\,\Delta P = m\Delta v \hfill \cr} \right.$

So, $m\Delta v \ge \sqrt {{h \over {4\pi }}} $

$\Delta v \ge \sqrt {{h \over {4\pi }}} \times {1 \over m}$

$\Delta v \ge {1 \over 2} > \sqrt {{h \over \pi }} > {1 \over m}$

$\Delta v \ge \sqrt {{h \over \pi }} {1 \over {2m}}$

Both the statements are true (correct). So answer is (2) Both statement I and statement II are true.

Bohr radius of hychogon atom $\rightarrow a_0$

According to Bohr, the equation used to calculate the angular momentum of an election in a hydrogen atom is

$\mathrm{mvr=\frac{nh}{2\pi}}$ ..... (1)

$m \rightarrow$ mass of electron

$v \rightarrow$ velocity of electron

$r \rightarrow$ radius of the orbit

$n \rightarrow$ orbit. number. in. which electron is present.

Given that; election is present in second orbit, $n=2$

The radius of the second orbit $r_2=a_0\times2^2=4a_0$

General formula for radius of $n^{th}$ orbit,

$r_n=a_0\times n^2$

From (1)

$\begin{aligned} & m v r=n \frac{h}{2 \pi} \\ & 2 \pi r=n \frac{h}{m v} \end{aligned}$

$\frac{h}{m v}=\lambda$ (de Broglie relation ship, $\lambda \rightarrow$ de Broglie Wavelength

So, $2 \pi r=n \lambda$

For the electron in the second orbit, $2 \pi r_2=n \lambda$

Substitute for $r_2$,

$\begin{aligned} & 2 \pi \times 4 a_0=n \lambda \\ & 8 \pi a_0=n \lambda \\ & \therefore \lambda=\frac{8 \pi a_0}{n} \end{aligned}$

Correct answer: Option 1) $\frac{8 \pi a_0}{n}$

First, recall the key fact that for a hydrogen atom (and hydrogen‐like ions), all orbitals having the same principal quantum number $n$ have the same energy. In other words,

$ E_{n\ell} \text{ depends only on }n\text{, and not on }\ell. $

Hence, the energy ordering for hydrogen‐like atoms follows

$ 1s \;<\; 2s = 2p \;<\; 3s = 3p = 3d \;<\; 4s = 4p = 4d = 4f \;<\;\dots $

Analyzing Each Statement

We check each statement against the true ordering for the hydrogen atom.

(A) $\;1s < 2p < 3d < 4s$

Because $2p$ is an $n=2$ orbital, $3d$ is $n=3$, and $4s$ is $n=4$, this partial ordering

$ E(1s) \;<\; E(2p) \;<\; E(3d)\;<\; E(4s) $

is consistent with the fact that $n=1 < n=2 < n=3 < n=4$.

Even though hydrogen also has 2s degenerate with 2p, and 3s and 3p degenerate with 3d, nothing in (A) contradicts the correct order $E_1 < E_2 < E_3 < E_4$.

So (A) is correct for hydrogen’s energy levels as stated.

(B) $\;1s < 2s = 2p < 3s = 3p$

For hydrogen, indeed $2s=2p$ in energy, and also $3s=3p=3d$.

The statement does not mention 3d, but it does correctly say $3s=3p$. That part is still true for hydrogen.

So for the orbitals it listed, (B) is also correct.

(C) $\;1s < 2s < 2p < 3s < 3p$

This is the kind of ordering that appears in many‐electron atoms (where subshell splitting occurs).

But for a hydrogen atom, $2s$ and $2p$ have exactly the same energy, so $2s<2p$ is not correct.

Hence (C) is not correct for hydrogen.

(D) $\;1s < 2s < 4s < 3d$

In multi‐electron atoms (e.g. the usual “(n+$\ell$) rule”), often $4s$ is lower than $3d$.

But for hydrogen, all orbitals with $n=3$ lie below all orbitals with $n=4$. That is, $3d$ is lower in energy than $4s$.

So (D) is not correct for hydrogen.

Conclusion

(C) and (D) are not correct for the hydrogen atom.

(A) and (B) are correct.

Hence the final answer matches the choice stating:

(C) and (D) only are not correct.

Looking at the given options, that is:

$ \boxed{\text{Option (D): (C) and (D) only.}} $

In absence of electric and magnetic fields, all orbitals of 3d are degenerate

In hydrogen atom the orbitals in a shell are degenerate means energy depends only on ' $n$ '

$\therefore \mathrm{E}_{3 \mathrm{p}_{\mathrm{x}}}=\mathrm{E}_{3 \mathrm{~d}_{x^2-y^2}}=\mathrm{E}_{3 \mathrm{~d}_{z^2}}$

For a shell total number of orbitals $=\mathrm{n}^2$

Magnetic quantum number have values $(-\ell$ to $+\ell$) including 0.

To determine which spectral line of the hydrogen atom suits the wavelength for heat treatment of muscular pain (900 nm), we begin by applying the Rydberg equation:

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

Given:

$\lambda = 900 \text{ nm} = 9 \times 10^{-5} \text{ cm}$

$\mathrm{R}_{\mathrm{H}} = 10^5 \text{ cm}^{-1}$

Hydrogen atom (Z = 1)

We proceed by solving the equation:

$ \frac{1}{\lambda \times \mathrm{R}_{\mathrm{H}}} = \frac{1}{n_1^2} - \frac{1}{n_2^2} $

Substituting the known values:

$ \frac{1}{9 \times 10^{-5} \text{ cm} \times 10^5 \text{ cm}^{-1}} = \frac{1}{n_1^2} - \frac{1}{n_2^2} $

This simplifies to:

$ \frac{1}{9} = \frac{1}{n_1^2} - \frac{1}{n_2^2} $

This condition is satisfied when $ n_1 = 3 $ and $ n_2 = \infty $.

Therefore, the appropriate spectral transition is from $ n_2 = \infty $ to $ n_1 = 3 $, which corresponds to the Paschen series transition $ \infty \rightarrow 3 $.

Let us analyze the two statements in the context of atomic orbitals and electronic transitions:

Statement (I)

“A spectral line will be observed for a $ 2 p_x \rightarrow 2 p_y $ transition.”

For an emission (or absorption) line to be observed, there must be a difference in energy between the initial and final states.

In a typical hydrogen-like or many-electron atom (without additional external fields or splitting effects), the three $2p$ orbitals ($2 p_x$, $2 p_y$, $2 p_z$) are degenerate—i.e., they all have the same energy.

Consequently, a transition from $2 p_x$ to $2 p_y$ (both having the same energy) would involve no energy change. Hence, no photon is emitted or absorbed for this “transition.” So you would not observe a spectral line for such a transition.

Therefore, Statement (I) is false.

Statement (II)

“$2 p_x$ and $2 p_y$ are degenerate orbitals.”

Orbitals within the same subshell (e.g., $2p$ subshell) are typically degenerate (same energy) in an isolated atom (especially a hydrogen-like atom).

Thus, $2 p_x$ and $2 p_y$ (and $2 p_z$) do indeed have the same energy.

Therefore, Statement (II) is true.

Conclusion

Statement (I): False

Statement (II): True

Hence, the correct choice is:

Option A: Statement I is false but Statement II is true.

$ r_n = \frac{n^2\, a_0}{Z} $

For the helium ion, which is hydrogen-like with a nuclear charge of $ Z = 2 $, the first excited state corresponds to $ n = 2 $. Substituting these values:

$ r_2 = \frac{(2)^2\, a_0}{2} = \frac{4\, a_0}{2} = 2\, a_0 $

Thus, the radius of the first excited state of the helium ion is $ 2\, a_0 $.

The definition of the candela involves a source emitting monochromatic radiation of a specific frequency and having a precise radiant intensity in a given direction. The candela is thereby defined based on both the frequency of the monochromatic light and its radiant intensity in watts per steradian. According to the given statement, we are dealing with a frequency identified as '$A$' $\times 10^{12}$ hertz and a radiant intensity specified as $\frac{1}{B}$ watt per steradian. However, there seems to be a small confusion in how the values are represented in the question. The correct interpretation and formulation based on the International System of Units (SI) are as follows:

The frequency of the monochromatic radiation that is used in defining the candela is 540 THz (tera hertz), so the value of $A$ should be 540. The specific radiant intensity mentioned relates to how the candela is defined in terms of a source's power output in a particular direction within a given frequency. The value that corresponds to the efficiency of human visual perception at this frequency is approximately 683 lumens per watt. Therefore, the number 683 is involved in specifying how many lumens correspond to a power of one watt of monochromatic light at 540 THz.

Given this context, the correct answer derives from understanding that:

• '$A$' represents the frequency in THz, which is 540 for the definition of the candela.
• '$B$' refers to the conversion factor or the luminous efficacy in lumens per watt at this frequency, which is 683 lumens per watt, not its reciprocal.

Therefore, the correct values for '$A$' and '$B$' are 540 and 683, respectively, making Option C the correct answer.

To compare the energies of electrons in a multielectron system, we must consider both the principal quantum number ($\mathrm{n}$) and the azimuthal quantum number ($\mathrm{l}$). In multielectron atoms, the energy levels are influenced by electron-electron repulsions, which modify the energy ordering compared to the hydrogen atom.

The general rule, called the "n+l" rule or Madelung rule, states that the energy of an electron in a multielectron atom is primarily determined by the sum of the principal quantum number ($\mathrm{n}$) and the azimuthal quantum number ($\mathrm{l}$). An orbital with a lower $(\mathrm{n} + \mathrm{l})$ value has lower energy. If two orbitals have the same $(\mathrm{n} + \mathrm{l})$ value, the orbital with the lower $\mathrm{n}$ value has lower energy.

Now, let's analyze each set of quantum numbers:

(A) $\mathrm{n}=4, \mathrm{l}=1$

$\mathrm{n} + \mathrm{l} = 4 + 1 = 5$

(B) $\mathrm{n}=4, \mathrm{l}=2$

$\mathrm{n} + \mathrm{l} = 4 + 2 = 6$

(C) $\mathrm{n}=3, \mathrm{l}=1$

$\mathrm{n} + \mathrm{l} = 3 + 1 = 4$

(D) $\mathrm{n}=3, \mathrm{l}=2$

$\mathrm{n} + \mathrm{l} = 3 + 2 = 5$

(E) $\mathrm{n}=4, \mathrm{l}=0$

$\mathrm{n} + \mathrm{l} = 4 + 0 = 4$

Based on the $\mathrm{n} + \mathrm{l}$ values, the order of increasing energy is:

(C) = (E) < (A) = (D) < (B)

Among (C) and (E), (C) has lower $\mathrm{n}$ value, so it is lower in energy:

(C) < (E) < (A) = (D) < (B)

Among (A) and (D), (D) has lower $\mathrm{n}$ value, so it is lower in energy:

Thus, the correct order is: (C) < (E) < (D) < (A) < (B)

The correct answer is Option C:

$ (\mathrm{C}) < (\mathrm{E}) < (\mathrm{D}) < (\mathrm{A}) < (\mathrm{B}) $

Matter consists of non-divisible atoms.

All the atoms of given element have same properties including mass.

Hence, statements (B) and (D) are incorrect.

According to Dalton’s theory, compounds are formed when atoms of different elements combine in fixed ratio.

The number of radial nodes in an orbital is given by the formula:

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

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

For a $3p$ orbital,

$n = 3$ (since it is the 3rd energy level),

$l = 1$ (since $p$ orbitals correspond to $l = 1$).

Let's use the formula to determine the number of radial nodes:

$ \text{Number of radial nodes} = 3 - 1 - 1 = 1 $

So, the correct answer is: Option C : 1

Option A: Nuclear charge $(\mathrm{z})$ is the correct answer.

Isoelectronic species are atoms and ions that have the same number of electrons. When comparing the sizes of isoelectronic species, the major factor that affects their size is the nuclear charge $(\mathrm{z})$, which is the charge of the nucleus or the number of protons in the nucleus. Here’s why:

The more protons there are in the nucleus (higher nuclear charge), the more positively charged the nucleus is, and thus the stronger it can attract the electrons towards it. This results in a smaller radius for the species. Conversely, if there are fewer protons in the nucleus, the attraction to the electrons is weaker, leading to a larger radius. Let's compare the species given:

• $\mathrm{F}^{-}$ (fluoride ion) has 9 protons in the nucleus and 10 electrons.
• $\mathrm{Ne}$ (neon atom) has 10 protons in the nucleus and 10 electrons.
• $\mathrm{Na}^{+}$ (sodium ion) has 11 protons in the nucleus and 10 electrons.

Despite having the same number of electrons, the $\mathrm{Na}^{+}$ ion will be the smallest among the three due to the higher nuclear charge of 11 protons, which attracts the same number of electrons more strongly than $\mathrm{Ne}$ with 10 protons and $\mathrm{F}^{-}$ with 9 protons. Hence, $\mathrm{Na}^{+}$ has a lower atomic radius compared to $\mathrm{Ne}$, which in turn is smaller than $\mathrm{F}^{-}$.

Considering Option B (size being the same), Option C (electron-electron interaction), and Option D (principal quantum number), none of these factors are as significant as the difference in nuclear charge for isoelectronic species.

The principal quantum number (n) is the same for all these isoelectronic species since they all have their outer electrons in the second energy level (n=2) in the ground state. Electron-electron interactions do contribute to energy levels and could affect the size slightly, but the primary and determining factor for isoelectronic species remains the nuclear charge, as it directly influences the effective nuclear charge felt by the electrons. Therefore, Option A is the best answer.

The four quantum numbers for the outermost electron of potassium can be determined by first understanding the electronic configuration of potassium (atomic number 19). The electron configuration is as follows:

$ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 $

The outermost electron resides in the 4s orbital.

The four quantum numbers are as follows:

Principal quantum number ($n$): This indicates the energy level or shell number of the electron and for the 4s orbital, $n = 4$.

Azimuthal quantum number ($l$): This indicates the subshell or shape of the orbital. For an $s$ orbital, $l = 0$.

Magnetic quantum number ($m$): This indicates the orientation of the orbital in space. Since there is only one orientation for $s$ orbitals, $m = 0$.

Spin quantum number ($s$): This can be either $+\frac{1}{2}$ or $-\frac{1}{2}$. Typically, the first electron in an orbital has a spin of $+\frac{1}{2}$.

Thus, the four quantum numbers for the outermost electron of potassium are:

$n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}$

This matches with Option B:

$\mathrm{n}=4, l=0, \mathrm{~m}=0, s=+\frac{1}{2}$

Degenerate Orbitals: Orbitals that have the same energy are called degenerate orbitals.

Hydrogen-like Atoms (Single-Electron Systems): In a hydrogen atom (or any one-electron system), the energy of an orbital depends only on the principal quantum number $n$, not on the azimuthal quantum number $l$. Therefore, all orbitals with the same $n$ (e.g., 3s, 3p, 3d) have the same energy and are degenerate.

Multi-electron Atoms: In atoms with more than one electron, the energy also depends on $l$ (and other factors like screening), so 3s, 3p, and 3d orbitals are not all degenerate in multi-electron atoms.

Now, let us evaluate the given statements:

Statement (I): “The orbitals having same energy are called as degenerate orbitals.”

This statement is true by definition.

Statement (II): “In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals.”

In a hydrogen atom, all orbitals with the same principal quantum number $n$ (like 3s, 3p, 3d) are degenerate.

Hence, this statement is false.

Therefore, the correct option is:

Option A: Statement I is true but Statement II is false.

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

Fact based.

The correct set of four quantum numbers for the valence electron of a rubidium atom ($Z = 37$) is:

Principal quantum number ($n$): This indicates the main energy level or shell of the electron. For rubidium, the valence electron is in the 5th energy level, so $n = 5$.

Azimuthal quantum number ($l$): This defines the subshell or orbital type. The possible values for $l$ range from 0 to $n-1$. For rubidium, the valence electron is in the $5s$ orbital, so $l = 0$.

Magnetic quantum number ($m_l$): This indicates the orientation of the orbital in space. For $l = 0$, the only possible value is $m_l = 0$.

Spin quantum number ($m_s$): This represents the spin of the electron, which can be either $+\frac{1}{2}$ or $-\frac{1}{2}$. Generally, when an electron is added to an unoccupied orbital, it starts with a spin of $+\frac{1}{2}$.

Therefore, the correct set of quantum numbers for the valence electron of rubidium is:

Option B: $5,0,0,+\frac{1}{2}$

Given are two statements :

• Assertion A : "In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface."
• Reason R : "When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place."

From the understanding of the photoelectric effect, we know that the light beam must have a frequency $v$ that is greater than the threshold frequency $v_0$ for electrons to be ejected from the metal surface. Therefore, Assertion A is correct.

However, Reason R is somewhat misleading in the context of the photoelectric effect. It's true that a photon transfers its energy to an electron when it strikes the electron. However, for the electron to be ejected (i.e., the photoelectric effect to occur), the energy of the photon, given by $E_{photon} = h \cdot v$, where $h$ is Planck's constant and $v$ is the frequency of the light, must be greater than the work function $\phi$ (the minimum energy needed to eject an electron) of the metal. This condition is not captured in Reason R.

Therefore, the correct answer is :

Option C : Assertion A is correct, but Reason R is not correct.

In Bohr's model, the angular momentum of an electron in an orbit is quantized, which can be represented as:

$mvr = n \cdot \frac{h}{2\pi}$

Where:

• (m) is the mass of the electron,
• (v) is the velocity of the electron,
• (r) is the radius of the nth orbit, and
• (n) is the principal quantum number ((n = 1) for the first orbit, (n = 2) for the second orbit, and so on).

The de Broglie wavelength ($\lambda$) is given by:

$\lambda = \frac{h}{mv}$

For the nth orbit, the radius (r) is $n^2$ times the radius of the first orbit ($r_0$, or $\alpha_0$ in your notation):

$r = n^2 \cdot \alpha_0$

The circumference of the nth orbit is $2\pi r$, which can be written as:

$2\pi r = n \cdot \lambda$

Substituting the expression for (r):

$2\pi \cdot n^2 \cdot \alpha_0 = n \cdot \lambda$

Solving for $\lambda$ gives:

$\lambda = 2\pi \cdot n \cdot \alpha_0$

For the 3rd orbit ((n = 3)), this simplifies to:

$\lambda = 6\pi \cdot \alpha_0$