Given below are two statements: Statement (I): It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle. Statement (II) If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is gesqrt(h)/(pi)×(1)/(2m)$gesqrt{\frac{h}{pi}}\times\frac{1}{2m}$ In the light of the above statements, choose the correct answer from the options given below:

### Related Formula (1)/(λ) = RZ²((1)/(n₁²) - (1)/(n₂²))$$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ ### Core Logic Statement I is a factual description of how the hydrogen emission spectrum is obtained. Dissociation yields excited atoms that emit light at discrete frequencies. So Statement I is true. For Statement II: First line of Lyman series for H atom (Z=1, n₁=1, n₂=2$Z=1, n_1=1, n_2=2$): (1)/(λ₁) = R(1)²((1)/(1²) - (1)/(2²)) = R(1 - (1)/(4)) = (3R)/(4)$$\frac{1}{\lambda_1} = R(1)^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$$ Second line of Balmer series for He^+$He^+$ (Z=2, n₁=2, n₂=4$Z=2, n_1=2, n_2=4$): (1)/(λ₂) = R(2)²((1)/(2²) - (1)/(4²)) = 4R((1)/(4) - (1)/(16)) = 4R((3)/(16)) = (3R)/(4)$$\frac{1}{\lambda_2} = R(2)^2\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4R\left(\frac{1}{4} - \frac{1}{16}\right) = 4R\left(\frac{3}{16}\right) = \frac{3R}{4}$$ Since (1)/(λ₁) = (1)/(λ₂)$\frac{1}{\lambda_1} = \frac{1}{\lambda_2}$, their wavelengths (and therefore frequencies) are exactly the same. Statement II is true. ### Step 1: Final Conclusion Both statements are correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Δ E = 13.6 Z² ((1)/(n₁²) - (1)/(n₂²)) eV$$\Delta E = 13.6 Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \text{ eV}$$ ### Core Logic Let's find the values of n₁$n_1$ and n₂$n_2$ for each transition: - (A) Paschen (1st$1^{st}$ line): n₁ = 3, n₂ = 4$n_1 = 3, n_2 = 4$ - (B) Balmer (2nd$2^{nd}$ line): n₁ = 2, n₂ = 4$n_1 = 2, n_2 = 4$ - (C) Paschen (3rd$3^{rd}$ line): n₁ = 3, n₂ = 6$n_1 = 3, n_2 = 6$ - (D) Bracket (4th$4^{th}$ line): n₁ = 4, n₂ = 8$n_1 = 4, n_2 = 8$ Calculating or comparing the energy values corresponding to these transitions yields the ascending order of energy. ### Step 1: Final Conclusion The correct ascending order of energy of the given lines is D < A < C < B$D < A < C < B$, corresponding to option (4). ### Pattern Recognition Sees: hydrogen spectral lines energy comparison. Trap: Confusing series limits with specific line numbers. Shortcut: Evaluate transition frequencies or wavelength gaps using Rydberg formula equivalents. ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Eₙ = -RH × Z²n²$$E_{n} = -R_{H} \times \frac{Z^{2}}{n^{2}}$$ Δ E = RH Z² ( (1)/(n₁²) - (1)/(n₂²) )$$\Delta E = R_{H} Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ ### Core Logic Given RH = 2.18 × 10⁻¹¹ ergs$R_{H} = 2.18 \times 10^{-11}\text{ ergs}$. Convert this to Joules: 1 Joule = 10⁷ ergs RH = 2.18 × 10⁻¹⁸ J$1\text{ Joule} = 10^{7}\text{ ergs} \implies R_{H} = 2.18 \times 10^{-18}\text{ J}$. Calculate energy difference per atom: Δ E = 2.18 × 10⁻¹⁸ × 1² [ 11² - 12² ]$$\Delta E = 2.18 \times 10^{-18} \times 1^{2} \left[ \frac{1}{1^{2}} - \frac{1}{2^{2}} \right]$$ Δ E = 2.18 × 10⁻¹⁸ × ( 1 - (1)/(4) ) = 2.18 × 10⁻¹⁸ × (3)/(4)$$\Delta E = 2.18 \times 10^{-18} \times \left( 1 - \frac{1}{4} \right) = 2.18 \times 10^{-18} \times \frac{3}{4}$$ Δ E = 1.635 × 10⁻¹⁸ Joule/atom$$\Delta E = 1.635 \times 10^{-18}\text{ Joule/atom}$$ ### Step 1: Conversion to per mole To find the energy per mole, multiply by Avogadro's number (NA = 6.02 × 10²³$N_A = 6.02 \times 10^{23}$): Δ Emole = 1.635 × 10⁻¹⁸ × 6.02 × 10²³ Joule/mole$$\Delta E_{mole} = 1.635 \times 10^{-18} \times 6.02 \times 10^{23}\text{ Joule/mole}$$ Δ Emole = 9.84 × 10⁵ Joule/mole ≈ 9.835 × 10⁵ J mol⁻¹$$\Delta E_{mole} = 9.84 \times 10^{5}\text{ Joule/mole} \approx 9.835 \times 10^{5}\text{ J mol}^{-1}$$ ### Pattern Recognition Energy gaps in Hydrogen: 1 arrow 2$1 \rightarrow 2$ transition is exactly (3)/(4)$\frac{3}{4}$ of the ionization energy. Watch out for per atom vs per mole unit traps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Δ E = 13.6 Z² ((1)/(n₁²) - (1)/(n₂²)) eV$$\Delta E = 13.6 Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\text{ eV}$$ ### Core Logic Step 1: First Balmer line (n₁ = 2, n₂ = 3$n_1 = 2, n_2 = 3$): Δ E₁ = x = 13.6 ((1)/(2²) - (1)/(3²)) = 13.6 ((1)/(4) - (1)/(9)) = 13.6 ((5)/(36))$$\Delta E_1 = x = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \left(\frac{5}{36}\right)$$ Step 2: Second Balmer line (n₁ = 2, n₂ = 4$n_1 = 2, n_2 = 4$): Δ E₂ = 13.6 ((1)/(2²) - (1)/(4²)) = 13.6 ((1)/(4) - (1)/(16)) = 13.6 ((3)/(16))$$\Delta E_2 = 13.6 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \left(\frac{3}{16}\right)$$ Step 3: Ratio of energies: (Δ E₂)/(x) = ((3)/(16))/((5)/(36)) = (3 × 36)/(16 × 5) = (27)/(20) = 1.35$$\frac{\Delta E_2}{x} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{3 \times 36}{16 \times 5} = \frac{27}{20} = 1.35$$ Δ E₂ = 1.35 x$$\Delta E_2 = 1.35 x$$ ### Pattern Recognition Sees: Ratio of hydrogen spectrum spectral line energies. Shortcut: (3/16) / (5/36) = 27/20 = 1.35$(3/16) / (5/36) = 27/20 = 1.35$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom Class 12 Physics: Atoms

### Related Formula Eₙ = -2.18 × 10⁻¹⁸ (Z²)/(n²) J/atom$$E_n = -2.18 \times 10^{-18} \frac{Z^2}{n^2} \text{ J/atom}$$ ### Core Logic Evaluate the energy of the stationary state for the given species by substituting the atomic number Z$Z$ and the orbit number n$n$ into the energy formula for hydrogen-like species. ### Step 1: Calculation for Li2+ Ion For the 3rd$3^{\text{rd}}$ orbit of the Li²⁺$Li^{2+}$ ion, we have Z = 3$Z = 3$ (since lithium has 3 protons) and n = 3$n = 3$. E₃ = -2.18 × 10⁻¹⁸ × (3²)/(3²)$$E_3 = -2.18 \times 10^{-18} \times \frac{3^2}{3^2}$$ E₃ = -2.18 × 10⁻¹⁸ J$$E_3 = -2.18 \times 10^{-18} \text{ J}$$ ### Pattern Recognition When Z = n$Z = n$, the Z²/n²$Z^2/n^2$ ratio becomes 1$1$, immediately resulting in the ground state energy of a hydrogen atom (-2.18 × 10⁻¹⁸$-2.18 \times 10^{-18}$ J). This is a common shortcut for identifying correct energy states. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom