For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n? [E: Energy of the stationary state, Z: atomic number, n: principal quantum number]

Related Formula

Bohr orbit radius expression varies with principle quantum number:

rₙ ∝ n²$r_n \propto n^2$

Core Logic

Let's check the proportional scaling across all choice items:

• Statement 1: (r₃)/(r₁) = (3²)/(1²) = 9$\frac{r_3}{r_1} = \frac{3^2}{1^2} = 9$ (Correct).
• Statement 2: (r₈)/(r₄) = (8²)/(4²) = (64)/(16) = 4$\frac{r_8}{r_4} = \frac{8^2}{4^2} = \frac{64}{16} = 4$ (Correct).
• Statement 3: (r₆)/(r₄) = ((6)/(4))² = (36)/(16) = (9)/(4) = 2.25 ≠ 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: (r₄)/(r₂) = ((4)/(2))² = 2² = 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 n²$n^2$ directly.

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$l$:

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$l=0$):

• For '2p' orbital (l=1$l=1$):

Step 1: Match the values

The orbital angular momentum values are 0$0$ and √(2)(h)/(2π)$\sqrt{2}\frac{h}{2\pi}$ 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$L=0$.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Related Formula

where, c = 3.0 × 10⁸ m s⁻¹$c = 3.0 \times 10^8 \text{ m s}^{-1}$ ν₀ = 5.16 × 10¹⁴ Hz$\nu_0 = 5.16 \times 10^{14} \text{ Hz}$

Core Logic

Let's first calculate the threshold wavelength (λ₀$\lambda_0$) for Caesium:

Comparing this with the visible spectrum:

• Yellow light (580 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 (620--750 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 (450--490 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 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.

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:

Let's systematically audit each statement:

• Statement A: Within the total population of 9 electrons, the pairing distribution across shells shows: 1s²$1s^2$ (one up, one down), 2s²$2s^2$ (one up, one down), 2p⁵$2p^5$ (three up, two down). Summing up-spins (mₛ = +(1)/(2)$m_s = +\frac{1}{2}$) yields 1 + 1 + 3 = 5$1 + 1 + 3 = 5$ electrons. Down-spins (mₛ = -(1)/(2)$m_s = -\frac{1}{2}$) 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 pₓ, py, pz$p_x, p_y, p_z$ 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 (pₓ, py$p_x, p_y$, or pz$p_z$) due to spatial symmetry. It is not constrained to pz$p_z$. 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.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Related Formula

Bohr energy level formula for hydrogenic species:

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 Be³⁺$\text{Be}^{3+}$:

• For Beryllium (Be$\text{Be}$), 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:

Step 3: Extract the magnitude and round to the nearest integer value:

Pattern Recognition

For the first excited state of Beryllium (Z=4, n=2$Z=4, n=2$), the term (Z²)/(n²) = (4²)/(2²) = (16)/(4) = 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 × 4 = 54.4$13.6 \times 4 = 54.4$).

Chapter Mix

Class 11 Chemistry: Structure of Atom