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.

To compare orbital energies, we analyze hydrogen (single-electron system) and lithium (multi-electron system) separately.

(i) Hydrogen atom (single-electron system):
In hydrogen-like species, the energy of an orbital depends only on the principal quantum number $ n $, not on $ l $. Therefore, all orbitals having the same $ n $ possess equal energy.

Hence, for hydrogen:

$ E_{2s}(H) = E_{2p}(H) $

So, Option B is correct.

(ii) Lithium atom (multi-electron system):
In multi-electron atoms, orbital energies depend on the $ (n + l) $ rule. The orbital with lower $ (n + l) $ value has lower energy. If two orbitals have the same $ n + l $, the one with lower $ n $ is lower in energy.

For lithium:

• For $ 2s $: $ n = 2, \, l = 0 \Rightarrow n + l = 2 $
• For $ 2p $: $ n = 2, \, l = 1 \Rightarrow n + l = 3 $

Since $ (n + l)_{2s} < (n + l)_{2p} $, we get:

$ E_{2s}(Li) < E_{2p}(Li) $

This is further supported by the fact that the $ 2s $ orbital penetrates closer to the nucleus and experiences higher effective nuclear charge, making it more stable (lower energy).

Thus, Option A is correct.

(iii) Comparison between hydrogen and lithium:
Energy of an orbital is given by:

$ E \propto -\frac{Z_{\text{eff}}^2}{n^2} $

In lithium, even after shielding by two inner $ 1s $ electrons, the effective nuclear charge experienced by the $ 2s $ electron is greater than that in hydrogen. As a result, the $ 2s $ electron in lithium is more strongly attracted and has lower (more negative) energy.

Therefore:

$ E_{2s}(Li) < E_{2s}(H) $

So, Option D is also correct.

(iv) Checking Option C:
Since $ E_{2p}(H) = E_{2s}(H) $ and $ E_{2s}(Li) $ is lower than $ E_{2s}(H) $, we have:

$ E_{2p}(H) > E_{2s}(Li) $

Thus, Option C is incorrect.

Final Answer: A, B and D

We know that the radius of any hydrogen-like species is given by the formula:

$ r_{\text{species}} = a_0 \frac{n^2}{Z} $

Here, $ a_0 $ is the Bohr radius of hydrogen (given as 52.9 pm), $ n $ is the principal quantum number, and $ Z $ is the atomic number of the nucleus.

Substituting the given values:

$ 70.53 = 52.9 \left( \frac{n^2}{Z} \right) $

Now, divide both sides by 52.9:

$ \frac{n^2}{Z} = \frac{70.53}{52.9} = 1.33 \approx \frac{4}{3} $

To make the fraction $ \frac{n^2}{Z} = \frac{4}{3} $, we can take $ n = 2 $ (so $ n^2 = 4 $) and $ Z = 3 $.

Thus, the species is $ \mathrm{Li}^{2+} $ and the electron is in the $ n = 2 $ energy level.

$ \mathrm{Li}^{2+},\ n = 2 \Rightarrow \frac{4}{3} = 1.33 $

For any hydrogen-like atom, the wavelength of emitted radiation is given by the Rydberg formula:

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

where

• $ R $ is the Rydberg constant,

• $ Z $ is the atomic number,

• $ n_1 $ and $ n_2 $ are integers with $ n_2 > n_1 $.

For hydrogen atom ($Z = 1$) — Lyman series

For the Lyman series, $ n_1 = 1 $ and $ n_2 = 2, 3, 4, \dots $

• The shortest wavelength corresponds to maximum energy difference, i.e. $ n_2 \to \infty $.

$ \frac{1}{x} = R (1)^2 \left( \frac{1}{1^2} - 0 \right) = R $

So,

$ x = \frac{1}{R} $

For helium ion, $\mathrm{He}^+$ ($Z = 2$) — Balmer series

For the Balmer series, $ n_1 = 2 $ and $ n_2 = 3, 4, 5, \dots $

• The longest wavelength corresponds to minimum energy difference, i.e. smallest $ n_2 $, which is $ n_2 = 3 $.

$ \frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) $

Simplify:

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

Thus,

$ \lambda = \frac{9}{5R} $

Express in terms of $x$

We have $ x = \frac{1}{R} $.

So,

$ \lambda = \frac{9}{5R} = \frac{9x}{5} $

✅ Final Answer

$ \boxed{\frac{9x}{5}} $

That is, Option A is correct.

A node is a region in space where the probability of finding an electron is zero. There are two types of nodes:

1. Radial Nodes (or Spherical Nodes): These are spherical surfaces where the probability of finding an electron is zero.

2. Angular Nodes (or Nodal Planes): These are planes (or cones for d-orbitals) passing through the nucleus where the probability of finding an electron is zero.

To find the number of these nodes for any given orbital, we use two simple formulas based on the principal quantum number ($n$) and the azimuthal quantum number ($l$).

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

• Number of angular nodes (nodal planes) = $l$

Remember, the value of $l$ depends on the subshell:

• For an 's' subshell, $l=0$

• For a 'p' subshell, $l=1$

• For a 'd' subshell, $l=2$

• For an 'f' subshell, $l=3$

Now, let's analyze each orbital from List-I.

A. 2s orbital

• Here, the principal quantum number $n = 2$.

• Since it is an 's' orbital, the azimuthal quantum number $l = 0$.

• Number of radial nodes = $n - l - 1 = 2 - 0 - 1 = 1$.

• Number of nodal planes = $l = 0$.

• So, the 2s orbital has 1 Radial node + No nodal plane. This matches with IV in List-II.

B. 3s orbital

• Here, $n = 3$.

• For an 's' orbital, $l = 0$.

• Number of radial nodes = $n - l - 1 = 3 - 0 - 1 = 2$.

• Number of nodal planes = $l = 0$.

• So, the 3s orbital has 2 Radial nodes + No nodal plane. This matches with III in List-II.

C. 3p orbital

• Here, $n = 3$.

• For a 'p' orbital, $l = 1$.

• Number of radial nodes = $n - l - 1 = 3 - 1 - 1 = 1$.

• Number of nodal planes = $l = 1$.

• So, the 3p orbital has 1 Radial node + one nodal plane. This matches with II in List-II.

D. 4d orbital

• Here, $n = 4$.

• For a 'd' orbital, $l = 2$.

• Number of radial nodes = $n - l - 1 = 4 - 2 - 1 = 1$.

• Number of nodal planes = $l = 2$.

• So, the 4d orbital has 1 Radial node + two nodal planes. This matches with I in List-II.

Let's summarize our findings:

• A matches with IV

• B matches with III

• C matches with II

• D matches with I

Comparing this with the given options, we find that the correct combination is A-IV, B-III, C-II, D-I.

Therefore, the correct answer is Option D.

Statement A: If two orbitals have the same value of $(n+l)$, the orbital with lower value of $n$ will have lower energy.

According to the Bohr-Bury rule (or the $(n+l)$ rule) for the filling of electrons in multi-electron atoms:

• Orbitals with a lower value of $(n+l)$ have lower energy.

• If two orbitals have the same $(n+l)$ value, the orbital with the lower principal quantum number ($n$) will have the lower energy.

For example, for the $3\mathrm{d}$ orbital, $(n+l) = 3+2 = 5$. For the $4\mathrm{p}$ orbital, $(n+l) = 4+1 = 5$. Since $3\mathrm{d}$ has a lower value of $n$ ($n=3$) compared to $4\mathrm{p}$ ($n=4$), the $3\mathrm{d}$ orbital has lower energy.

Therefore, Statement A is true.

Statement B: Energies of the orbitals in the same subshell increase with increase in atomic number.

As the atomic number ($Z$) increases, the nuclear charge increases. This leads to a stronger electrostatic pull on the electrons by the nucleus, which increases the effective nuclear charge ($Z_{\mathrm{eff}}$). Due to this stronger attraction, the electrons are held more tightly, and the energy of the orbitals becomes more negative (it decreases, not increases). For example, the energy of a $2\mathrm{s}$ orbital follows the order: $E_{2\mathrm{s}}(\mathrm{H}) > E_{2\mathrm{s}}(\mathrm{Li}) > E_{2\mathrm{s}}(\mathrm{Na})$.

Therefore, Statement B is false.

Statement C: The size of $2\mathrm{p}_x$ orbital is less than the size of $3\mathrm{p}_x$ orbital.

The size of an orbital is primarily determined by its principal quantum number ($n$). As the value of $n$ increases, the distance of the electron from the nucleus increases, making the orbital larger. Since $n=3$ for the $3\mathrm{p}_x$ orbital and $n=2$ for the $2\mathrm{p}_x$ orbital, the $3\mathrm{p}_x$ orbital is larger than the $2\mathrm{p}_x$ orbital.

Therefore, Statement C is true.

Statement D: Among $5\mathrm{f}, 6\mathrm{s}, 4\mathrm{d}, 5\mathrm{p}$ and $5\mathrm{d}$ orbitals, none of the orbitals have 2 radial nodes.

The number of radial nodes for any orbital is calculated using the formula: $(n - l - 1)$.

Let us calculate the radial nodes for the given orbitals:

• For $5\mathrm{f}$ ($n=5, l=3$): Radial nodes $= 5 - 3 - 1 = 1$

• For $6\mathrm{s}$ ($n=6, l=0$): Radial nodes $= 6 - 0 - 1 = 5$

• For $4\mathrm{d}$ ($n=4, l=2$): Radial nodes $= 4 - 2 - 1 = 1$

• For $5\mathrm{p}$ ($n=5, l=1$): Radial nodes $= 5 - 1 - 1 = 3$

• For $5\mathrm{d}$ ($n=5, l=2$): Radial nodes $= 5 - 2 - 1 = 2$

As we can see, the $5\mathrm{d}$ orbital indeed has exactly 2 radial nodes. Thus, the statement that none of them has 2 radial nodes is incorrect.

Therefore, Statement D is false.

Conclusion:

Only statements A and C are true.

The correct answer is Option B.

For multi-electron atoms, the energy of an orbital is decided by the $n + l$ rule.

• The orbital with smaller value of $n+l$ has lower energy.

• If two orbitals have the same value of $n+l$, then the orbital with smaller $n$ has lower energy.

Also, energy does not depend on $m$.

Now let us identify each orbital:

• A: $n=3,\; l=2$

  This is $3d$

  $n+l=3+2=5$

• B: $n=4,\; l=0$

  This is $4s$

  $n+l=4+0=4$

• C: $n=6,\; l=1$

  This is $6p$

  $n+l=6+1=7$

• D: $n=5,\; l=1$

  This is $5p$

  $n+l=5+1=6$

• E: $n=2,\; l=1$

  This is $2p$

  $n+l=2+1=3$

Now arrange in increasing energy:

$ 2p < 4s < 3d < 5p < 6p $

So,

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

Hence, the correct answer is:

$ \boxed{\text{Option D}} $

Statement A

Heisenberg uncertainty principle is applicable to electrons.

This statement is correct.

According to Heisenberg uncertainty principle, it is impossible to determine simultaneously the exact position and exact momentum of a microscopic particle like an electron.

So, A is true.

Statement B

The size of $2p_x$ orbital is less than the size of $3p_x$ orbital.

This statement is correct.

As the principal quantum number $n$ increases, the size of the orbital increases.

Here,

• $2p_x$ has $n = 2$

• $3p_x$ has $n = 3$

Therefore, size of $3p_x$ orbital is greater than that of $2p_x$ orbital.

So, B is true.

Statement C

The energy of $2s$ orbital of H atom is equal to the energy of $2s$ orbital of Li.

This statement is incorrect.

For hydrogen atom, energy depends only on principal quantum number $n$. But for multi-electron atoms like lithium, the energy depends on both $n$ and $l$ due to shielding and penetration effects.

So, the energy of $2s$ orbital in H and Li is not equal.

Thus, C is false.

Statement D

The electronic configuration of Cr is $[\mathrm{Ar}]\,3d^5\,4s^1$

This statement is correct.

Chromium is an exception to the expected configuration. Due to extra stability of half-filled $3d$ subshell, its configuration is:

$ [\mathrm{Ar}]\,3d^5\,4s^1 $

So, D is true.

Final Answer

Correct statements are A, B and D only.

Therefore, the correct option is:

$ \boxed{\text{Option B}} $

For a hydrogen-like species, according to Bohr’s theory, the radius of the $n^{\text{th}}$ orbit is

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

where:

• $n$ = orbit number

• $Z$ = atomic number

• $a_0$ = Bohr radius

So, for identical radii, the value of $\dfrac{n^2}{Z}$ must be same.

Now check each species:

A. H (first orbit)

For hydrogen, $Z=1$, $n=1$

$ r = \frac{1^2 a_0}{1} = a_0 $

B. $\mathrm{He}^{+}$ (first orbit)

Here, $Z=2$, $n=1$

$ r = \frac{1^2 a_0}{2} = \frac{a_0}{2} $

C. $\mathrm{He}^{+}$ (second orbit)

Here, $Z=2$, $n=2$

$ r = \frac{2^2 a_0}{2} = \frac{4a_0}{2} = 2a_0 $

D. $\mathrm{Li}^{2+}$ (first orbit)

Here, $Z=3$, $n=1$

$ r = \frac{1^2 a_0}{3} = \frac{a_0}{3} $

E. $\mathrm{Be}^{3+}$ (second orbit)

Here, $Z=4$, $n=2$

$ r = \frac{2^2 a_0}{4} = \frac{4a_0}{4} = a_0 $

Now compare:

• A: $a_0$

• B: $\dfrac{a_0}{2}$

• C: $2a_0$

• D: $\dfrac{a_0}{3}$

• E: $a_0$

So, the identical radii are for A and E.

Therefore, the correct answer is:

$ \boxed{\text{Option B: A and E Only}} $

For hydrogen spectrum, the wave number is given by the Rydberg formula:

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

where:

• $n_1$ = lower energy level

• $n_2$ = higher energy level

• $n_2 > n_1$

Now let us find the first line of each series.

1. First line of Balmer series

For Balmer series:

$ n_1 = 2 $

The first line means transition from $n_2 = 3$ to $n_1 = 2$.

So,

$ \bar{\nu}_B = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) $

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

2. First line of Brackett series

For Brackett series:

$ n_1 = 4 $

The first line means transition from $n_2 = 5$ to $n_1 = 4$.

So,

$ \bar{\nu}_{Br} = R\left(\frac{1}{4^2} - \frac{1}{5^2}\right) = R\left(\frac{1}{16} - \frac{1}{25}\right) $

$ \bar{\nu}_{Br} = R\left(\frac{25-16}{400}\right) = \frac{9R}{400} $

3. Required ratio

$ \bar{\nu}_B : \bar{\nu}_{Br} = \frac{5R}{36} : \frac{9R}{400} $

Cancel $R$:

$ = \frac{5}{36} : \frac{9}{400} $

$ = \frac{5}{36} \times \frac{400}{9} $

So ratio:

$ \bar{\nu}_B : \bar{\nu}_{Br} = \frac{2000}{324} = \frac{500}{81} $

Now,

$ \frac{500}{81} = \frac{5 \times 100}{81} $

and

$ \frac{100}{81} \approx 1.23 $

So,

$ \bar{\nu}_B : \bar{\nu}_{Br} = 5 : 0.81 $

Therefore, the correct answer is:

$ \boxed{\text{Option B } \; 5 : 0.81} $

Using the Aufbau principle, we fill electrons in the order:

$ 1s,\ 2s,\ 2p,\ 3s,\ 3p,\ 4s,\ 3d,\dots $

Now let us count electrons one by one up to the $19^{\text{th}}$ electron.

• $1s^2 \Rightarrow 2$ electrons

• $2s^2 \Rightarrow 4$ electrons

• $2p^6 \Rightarrow 10$ electrons

• $3s^2 \Rightarrow 12$ electrons

• $3p^6 \Rightarrow 18$ electrons

So, the $19^{\text{th}}$ electron enters the $4s$ orbital.

For a $4s$ electron:

• $n=4$

• for $s$-orbital, $l=0$

• when $l=0$, only possible value of $m$ is $0$

• for the first electron in this orbital, $s=+\dfrac{1}{2}$

Thus, the set of four quantum numbers is:

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

So, the correct option is:

$ \boxed{\text{Option D}} $

Note: Chromium actually has exceptional configuration $[Ar]\,3d^5 4s^1$, but the question clearly says "in accordance with Aufbau principle", so we must place the $19^{\text{th}}$ electron in $4s$.

For the process

$ \text{Li}^{2+}(g) \rightarrow \text{Li}^{3+}(g) + e^- $

we have to remove the only electron present in $\text{Li}^{2+}$.

$\text{Li}^{2+}$ is a hydrogen-like species because it has only one electron.

For a hydrogen-like species, ionization energy is given by

$ E = Z^2 \times 2.18 \times 10^{-18}\ \text{J atom}^{-1} $

where $Z$ is the atomic number.

For lithium, $Z = 3$.

So,

$ E = 3^2 \times 2.18 \times 10^{-18} $

$ E = 9 \times 2.18 \times 10^{-18} $

$ E = 19.62 \times 10^{-18} $

$ E = 1.962 \times 10^{-17}\ \text{J atom}^{-1} $

Hence, the required energy is

$ \boxed{1.962 \times 10^{-17}\ \text{J atom}^{-1}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

Atomic number $9$ means the element has $9$ electrons. Its electronic configuration is

$ 1s^2\,2s^2\,2p^5 $

The orbital arrangement can be shown as

$ 1s:\uparrow\downarrow \qquad 2s:\uparrow\downarrow \qquad 2p:\boxed{\uparrow\downarrow}\ \boxed{\uparrow\downarrow}\ \boxed{\uparrow} $

For statement A:

In $1s^2$ and $2s^2$, each pair has one electron with $m_s=+\frac{1}{2}$ and one with $m_s=-\frac{1}{2}$.

In $2p^5$, according to Hund’s rule, there are $3$ electrons with one spin direction and $2$ with the opposite spin direction.

So total can be:

$ m_s=+\frac{1}{2}: 5 $

$ m_s=-\frac{1}{2}: 4 $

So A is correct.

For statement B:

In $2p^5$, there is one unpaired electron in one of the three $p$ orbitals. But we cannot specifically say it must be in $p_z$. The $p_x$, $p_y$, and $p_z$ orbitals are degenerate, so the singly occupied orbital is not fixed as $p_z$.

So B is not correct.

For statement C:

The last electron enters the $2p$ subshell.

For a $2p$ electron,

$ n=2,\qquad l=1 $

So C is correct.

For statement D:

Number of angular nodes is equal to $l$.

The occupied atomic orbitals are $1s$, $2s$, and the three $2p$ orbitals.

$ 1s: l=0 $

$ 2s: l=0 $

$ 2p_x, 2p_y, 2p_z: l=1 \text{ each} $

So the sum of angular nodes of the occupied atomic orbitals is

$ 0+0+1+1+1=3 $

not $1$.

So D is not correct.

The correct statements are A and C only.

$ \boxed{\text{Option C}} $

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.