### Related Formula Bohr energy level formula for hydrogenic species: E_n = -13.6 times fracZ^2n^2 quad texteV$E_n = -13.6 \times \frac{Z^2}{n^2} \quad \text{eV}$ where: Z$Z$ = atomic number of the species n$n$ = principal quantum number of the orbit ### Execution Step 1: Identify the parameters for the first excited state of textBe^3+$\text{Be}^{3+}$: * For Beryllium (textBe$\text{Be}$), the atomic number is Z = 4$Z = 4$. * The term 'first excited state' refers to the second energy level, so n = 2$n = 2$. Step 2: Substitute these values into the Bohr energy equation: E_textBe^3+ = -13.6 times frac4^22^2 = -13.6 times frac164$E_{\text{Be}^{3+}} = -13.6 \times \frac{4^2}{2^2} = -13.6 \times \frac{16}{4}$ E_textBe^3+ = -13.6 times 4 = -54.4 text eV$E_{\text{Be}^{3+}} = -13.6 \times 4 = -54.4 \text{ eV}$ Step 3: Extract the magnitude and round to the nearest integer value: |E_textBe^3+| = 54.4 approx 54$|E_{\text{Be}^{3+}}| = 54.4 \approx 54$ ### Pattern Recognition For the first excited state of Beryllium (Z=4, n=2$Z=4, n=2$), the term fracZ^2n^2 = frac4^22^2 = frac164 = 4$\frac{Z^2}{n^2} = \frac{4^2}{2^2} = \frac{16}{4} = 4$. Thus, the energy value is exactly 4 times that of the ground-state hydrogen atom (13.6 times 4 = 54.4$13.6 \times 4 = 54.4$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula E_n = -13.6 fracZ^2n^2text eV$E_n = -13.6 \frac{Z^2}{n^2}\text{ eV}$ ### Core Logic For a constant principal quantum number n$n$, the energy E$E$ is directly proportional to -Z^2$-Z^2$. This represents a quadratic relation where the curve is a downward-opening parabola starting from the origin in the negative energy region. ### Pattern Recognition Since energy values are inherently negative for bound states, as Z$Z$ increases, E$E$ becomes rapidly more negative following a parabolic curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Delta x cdot Delta p ge frach4pi$Delta x cdot Delta p ge \frac{h}{4pi}$ ### Core Logic Statement I is a verbatim definition of Heisenberg's Uncertainty Principle, hence it is completely true. For Statement II, we are given that Delta x = Delta p$Delta x = Delta p$: Delta p cdot Delta p ge frach4pi implies (Delta p)^2 ge frach4pi $Delta p cdot Delta p ge \frac{h}{4pi} implies (Delta p)^2 ge \frac{h}{4pi} $ Delta p ge sqrtfrach4pi = frac12sqrtfrachpi$Delta p ge sqrt{\frac{h}{4pi}} = \frac{1}{2}sqrt{\frac{h}{pi}}$ Since Delta p = m cdot Delta v$Delta p = m cdot Delta v$: m cdot Delta v ge frac12sqrtfrachpi implies Delta v ge frac12msqrtfrachpi$m cdot Delta v ge \frac{1}{2}sqrt{\frac{h}{pi}} implies Delta v ge \frac{1}{2m}sqrt{\frac{h}{pi}}$ This perfectly matches Statement II, so it is also true. ### Pattern Recognition When solving inequality bounds for identical uncertainties, always substitute directly to obtain a clean quadratic form before taking the square root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Core Logic In multi-electron systems, the energy profile depends on both the primary shell (n$n$) and azimuthal subshell (l$l$) quantum values via the (n+l)$(n+l)$ rule. Orbitals sharing the exact same n$n$ and l$l$ values are energy-degenerate as long as external field metrics are zero. Let us map each set: - **A:** 1s$1s$ - **B:** 2s$2s$ - **C:** 2p$2p$ - **D:** 3d$3d$ (m_l = 1$m_l = 1$) - **E:** 3d$3d$ (m_l = 0$m_l = 0$) Since both D and E describe subcomponents of the same 3d$3d$ subshell (n=3, l=2$n=3, l=2$), they share identical energies, establishing perfect degeneracy. ### Pattern Recognition Sees: Orbitals matching energy level in multi-electron space. Trap: Confusing single-electron hydrogen atoms (where energy depends solely on n$n$) with multielectron atoms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula According to the quantum mechanical model, the position and momentum of an electron cannot be determined simultaneously with absolute certainty as per Heisenberg's Uncertainty Principle: Delta x cdot Delta p ge frach4pi $\Delta x \cdot \Delta p \ge \frac{h}{4\pi} $ ### Core Logic Bohr's model defines fixed circular orbits for electrons around the nucleus, implying precise knowledge of both trajectory and position. However, the quantum mechanical model replaces these deterministic circular paths with three-dimensional probability distributions known as orbitals, where the path is spherical for the ground state of hydrogen. ### Step 1: Evaluation of Options Postulates (1), (2), and (3) regarding discrete energy levels, stationary states, and photon emissions are preserved in the quantum mechanical framework. Postulate (4) incorrectly confines the electron to a defined circular two-dimensional path, which directly contradicts quantum mechanical principles. ### Pattern Recognition Sees: "Bohr's model not in agreement with Quantum Mechanics" ightarrow$ ightarrow$ Look for deterministic trajectories like "moves in a circle" or "fixed path". Quantum mechanics is probabilistic, not deterministic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom