### Related Formula textNumber of spectral lines = fracDelta n (Delta n + 1)2$\text{Number of spectral lines} = \frac{\Delta n (\Delta n + 1)}{2}$ where Delta n = n_2 - n_1$\Delta n = n_2 - n_1$ ### Core Logic Identify the principal quantum numbers (n$n$) for the given states: - Fifth excited state means n_2 = 5 + 1 = 6$n_2 = 5 + 1 = 6$. - First excited state means n_1 = 1 + 1 = 2$n_1 = 1 + 1 = 2$. Calculate the difference: Delta n = n_2 - n_1 = 6 - 2 = 4$\Delta n = n_2 - n_1 = 6 - 2 = 4$ ### Step 1: Calculate Total Spectral Lines Substitute Delta n = 4$\Delta n = 4$ into the formula: textMaximum number of spectral lines = frac4(4 + 1)2$\text{Maximum number of spectral lines} = \frac{4(4 + 1)}{2}$ = frac4 times 52$= \frac{4 \times 5}{2}$ = 10$= 10$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Core Logic Statement-I defines degenerate orbitals correctly: Orbitals that share the exact same energy level are called degenerate orbitals. Statement-II claims 3p and 3d are not degenerate in a hydrogen atom. For a single-electron system like Hydrogen (1s^1$1s^1$), the energy of an orbital depends *only* on the principal quantum number 'n$n$'. ### Step 1: Final conclusion In a hydrogen atom, energy of 3s = 3p = 3d$3s = 3p = 3d$. Therefore, 3p and 3d *are* degenerate. Thus, Statement-II is false. ### Pattern Recognition Hydrogen atom (single electron species) = Energy dictated strictly by n$n$. Multi-electron atoms = Energy dictated by (n + l)$(n + l)$ rule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Core Logic The atomic number of potassium (K) is 19$19$. Its electronic configuration is: 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$. The outermost orbital of potassium is the 4s$4s$ orbital. For the 4s$4s$ orbital: Principal quantum number, n = 4$n = 4$ Azimuthal quantum number, l = 0$l = 0$ (for s-subshell) Magnetic quantum number, m = 0$m = 0$ (since l = 0$l = 0$) Spin quantum number, s = pm frac12$s = \pm \frac{1}{2}$ ### Step 1: Final Matching Comparing these values with the options, option (2) provides n = 4, l = 0, m = 0, s = +frac12$n = 4, l = 0, m = 0, s = +\frac{1}{2}$, which perfectly describes the outermost electron. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula E = frac1240lambda (textnm) text eV$E = \frac{1240}{\lambda (\text{nm})} \text{ eV}$ E_texttotal = E times N_A$E_{\text{total}} = E \times N_A$ ### Step 1: Energy per atom E = frac1240242 text eV = 5.12 text eV$E = \frac{1240}{242} \text{ eV} = 5.12 \text{ eV}$ Converting eV to Joules: E = 5.12 times 1.6 times 10^-19 text J/atom$E = 5.12 \times 1.6 \times 10^{-19} \text{ J/atom}$ E = 8.192 times 10^-19 text J/atom$E = 8.192 \times 10^{-19} \text{ J/atom}$ ### Step 2: Energy per mole To find the ionization energy per mole, multiply by Avogadro's number (N_A approx 6.022 times 10^23$N_A \approx 6.022 \times 10^{23}$): IE = 8.192 times 10^-19 times 6.022 times 10^23 text J/mol$IE = 8.192 \times 10^{-19} \times 6.022 \times 10^{23} \text{ J/mol}$ IE approx 493.3 times 10^3 text J/mol approx 494 text kJ/mol$IE \approx 493.3 \times 10^3 \text{ J/mol} \approx 494 \text{ kJ/mol}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom