### Related Formula psi_n,l,m(r, theta, phi) = R_n,l(r) cdot Y_l,m(theta, phi)$\psi_{n,l,m}(r, \theta, \phi) = R_{n,l}(r) \cdot Y_{l,m}(\theta, \phi)$ ### Core Logic Let us evaluate each statement carefully: - **Statement (A)** is **True**: The subsidiary (or azimuthal) quantum number l$l$ dictates the orbital shape (l=0 implies textspherical s$l=0 \implies \text{spherical } s$, l=1 implies textdumbbell p$l=1 \implies \text{dumbbell } p$, l=2 implies textdouble-dumbbell d$l=2 \implies \text{double-dumbbell } d$, etc.).
- **Statement (B)** is **True**:

The diagram clearly displays a boundary surface showing two symmetrical lobes situated along the x-axis, which is the exact depiction of a 2mathrmp_x$2\mathrm{p_x}$ orbital.
- **Statement (C)** is **False**: The +$+$ and -$-$ signs in orbital diagrams indicate the mathematical sign (or phase) of the spatial wave function psi$\psi$ in those regions, not electric charge.
- **Statement (D)** is **False**: The wave function of a 2mathrmp_x$2\mathrm{p_x}$ orbital is zero everywhere in its nodal plane. For a 2mathrmp_x$2\mathrm{p_x}$ orbital, the nodal plane is the **yz-plane** (i.e., at x=0$x=0$), not the xy-plane. ### Step 1: Conclusion Only statements (A) and (B) are correct. ### Pattern Recognition Nodal planes of p-orbitals are perpendicular to the orbital axis: - p_x rightarrow yz$p_x \rightarrow yz$-plane is node (x=0$x=0$) - p_y rightarrow xz$p_y \rightarrow xz$-plane is node (y=0$y=0$) - p_z rightarrow xy$p_z \rightarrow xy$-plane is node (z=0$z=0$) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Bohr orbit radius expression varies with principle quantum number: r_n propto n^2$r_n \propto n^2$ ### Core Logic Let's check the proportional scaling across all choice items: * Statement 1: fracr_3r_1 = frac3^21^2 = 9$\frac{r_3}{r_1} = \frac{3^2}{1^2} = 9$ (Correct). * Statement 2: fracr_8r_4 = frac8^24^2 = frac6416 = 4$\frac{r_8}{r_4} = \frac{8^2}{4^2} = \frac{64}{16} = 4$ (Correct). * Statement 3: fracr_6r_4 = left(frac64right)^2 = frac3616 = frac94 = 2.25 neq 3$\frac{r_6}{r_4} = \left(\frac{6}{4}\right)^2 = \frac{36}{16} = \frac{9}{4} = 2.25 \neq 3$ (Incorrect statement). * Statement 4: fracr_4r_2 = left(frac42right)^2 = 2^2 = 4$\frac{r_4}{r_2} = \left(\frac{4}{2}\right)^2 = 2^2 = 4$ (Correct). ### Step 1: Identification Therefore, choice (3) is mathematically incorrect. ### Pattern Recognition Remember to square the shell indexes immediately! Don't look at the linear ratio (6/4 = 1.5$6/4 = 1.5$), always apply the quadratic scaling factor n^2$n^2$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom Class 12 Physics: Atoms

### Related Formula Orbital angular momentum (L$L$) of an electron is determined exclusively by its azimuthal quantum number l$l$: L = sqrtl(l+1) frach2pi$L = \sqrt{l(l+1)} \frac{h}{2\pi}$ where values of l$l$ are: - For s-orbitals: l=0$l=0$ - For p-orbitals: l=1$l=1$ - For d-orbitals: l=2$l=2$ ### Core Logic Evaluate for both specified orbitals: - For '2s' orbital (l=0$l=0$): L = sqrt0(0+1) frach2pi = 0$L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$ - For '2p' orbital (l=1$l=1$): L = sqrt1(1+1) frach2pi = sqrt2 frach2pi$L = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$ ### Step 1: Match the values The orbital angular momentum values are 0$0$ and sqrt2frach2pi$\sqrt{2}\frac{h}{2\pi}$ respectively, corresponding to Option (4). ### Pattern Recognition Always note that orbital angular momentum does *not* depend on the principal quantum number n$n$ (the '2' in 2s and 2p is irrelevant). An electron in a 1s, 2s, or 3s orbital always has L=0$L=0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula lambda = fraccnu$\lambda = \frac{c}{\nu}$ where, c = 3.0 times 10^8 text m s^-1$c = 3.0 \times 10^8 \text{ m s}^{-1}$ nu_0 = 5.16 times 10^14 text Hz$\nu_0 = 5.16 \times 10^{14} \text{ Hz}$ ### Core Logic Let's first calculate the threshold wavelength (lambda_0$\lambda_0$) for Caesium: lambda_0 = frac3.0 times 10^85.16 times 10^14 approx 5.81 times 10^-7 text m = 581.4 text nm$\lambda_0 = \frac{3.0 \times 10^8}{5.16 \times 10^{14}} \approx 5.81 \times 10^{-7} \text{ m} = 581.4 \text{ nm}$ Comparing this with the visible spectrum: - Yellow light (sim 580 text nm$\sim 580 \text{ nm}$) corresponds closely to the threshold wavelength. Hence, photoelectric emission occurs (Statement A is correct). - The photocurrent is directly proportional to the light intensity (brightness). Reducing brightness decreases the current (Statement B is correct). - Red light (sim 620text--750 text nm$\sim 620\text{--}750 \text{ nm}$) has a lower frequency than the threshold frequency, meaning no current is produced (Statement C is incorrect). - Blue light (sim 450text--490 text nm$\sim 450\text{--}490 \text{ nm}$) has a higher frequency than the threshold. Thus, current is produced (Statement D is correct). - White light contains all wavelengths (including blue, violet, etc.), which will produce a photoelectric current (Statement E is correct). Therefore, statements A, B, D, and E are correct. ### Pattern Recognition The wavelength threshold is 581 text nm$581 \text{ nm}$ (yellow-green region). Any light with shorter wavelength (higher frequency), like blue or white, will cause emission. Dimming simply decreases photon flux and hence current. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom Class 12 Physics: Dual Nature of Radiation and Matter

### Core Logic The element with atomic number 9 is **Fluorine (F)**. Let's write out its ground state electronic configuration: F (Z=9) = 1s^2 \, 2s^2 \, 2p^5$F (Z=9) = 1s^2 \, 2s^2 \, 2p^5$ Let's systematically audit each statement: * **Statement A**: Within the total population of 9 electrons, the pairing distribution across shells shows: 1s^2$1s^2$ (one up, one down), 2s^2$2s^2$ (one up, one down), 2p^5$2p^5$ (three up, two down). Summing up-spins (m_s = +frac12$m_s = +\frac{1}{2}$) yields 1 + 1 + 3 = 5$1 + 1 + 3 = 5$ electrons. Down-spins (m_s = -frac12$m_s = -\frac{1}{2}$) yield 1 + 1 + 2 = 4$1 + 1 + 2 = 4$ electrons. **Statement A is fully correct.**

* **Statement B**: By Hund's Rule, the 5 electrons in the 2p subshell occupy the degenerate p_x, p_y, p_z$p_x, p_y, p_z$ states. This produces two fully-filled sub-orbitals and one half-filled sub-orbital. The unpaired slot can reside arbitrarily in *any* of the three orbitals (p_x, p_y$p_x, p_y$, or p_z$p_z$) due to spatial symmetry. It is not constrained to p_z$p_z$. **Statement B is incorrect.** * **Statement C**: The highest energy valence electron enters the 2p subshell, which is defined by principal number n = 2$n = 2$ and azimuthal index l = 1$l = 1$. **Statement C is fully correct.** * **Statement D**: Angular nodes are given directly by the quantum number l$l$. For s-orbitals (1s, 2s$1s, 2s$), angular nodes = 0$= 0$. For each of the three populated p-orbitals (2p$2p$), angular nodes = 1$= 1$. The sum total of angular nodes across all orbitals is 0 + 0 + 3 = 3$0 + 0 + 3 = 3$. **Statement D is incorrect.** ### Pattern Recognition Total angular nodes equals the total number of p-electrons' spatial orientation count, not simply the subshell boundary value. Recognizing that degenerate p-orbitals share uniform probability status exposes the restriction in Statement B instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom