Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 times 10^5 text m/s$3 \times 10^{5} \text{ m/s}$ and 2.5 times 10^5 text m/s$2.5 \times 10^{5} \text{ m/s}$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are ____ respectively.

### Related Formula K_textinitial = U_textclosest approach$$K_{\text{initial}} = U_{\text{closest approach}}$$ K = frac14pi epsilon_0 frac(2e)(Ze)r_0$$K = \frac{1}{4\pi \epsilon_{0}} \frac{(2e)(Ze)}{r_0}$$ ### Core Logic By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r_0$r_0$). K_i + U_i = K_f + U_f$K_i + U_i = K_f + U_f$ K_i + 0 = 0 + frac14piepsilon_0 frac(2e)(79e)r_0$$K_i + 0 = 0 + \frac{1}{4\pi\epsilon_0} \frac{(2e)(79e)}{r_0}$$ ### Step 1: Convert Energy and Solve Initial kinetic energy K_i = 7.7text MeV = 7.7 times 10^6 times 1.6 times 10^-19text J$K_i = 7.7\text{ MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19}\text{ J}$. 7.7 times 10^6 times 1.6 times 10^-19 = frac9 times 10^9 times (2 times 1.6 times 10^-19) times (79 times 1.6 times 10^-19)r_0$$7.7 \times 10^6 \times 1.6 \times 10^{-19} = \frac{9 \times 10^9 \times (2 \times 1.6 \times 10^{-19}) \times (79 \times 1.6 \times 10^{-19})}{r_0}$$ r_0 = frac9 times 10^9 times 2 times 79 times (1.6 times 10^-19)^27.7 times 10^6 times 1.6 times 10^-19$$r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}$$ r_0 = frac9 times 10^9 times 158 times 1.6 times 10^-197.7 times 10^6$$r_0 = \frac{9 \times 10^9 \times 158 \times 1.6 \times 10^{-19}}{7.7 \times 10^6}$$ r_0 = frac2275.2 times 10^-107.7 times 10^6 = 295.48 times 10^-16text m approx 2.95 times 10^-14text m$$r_0 = \frac{2275.2 \times 10^{-10}}{7.7 \times 10^6} = 295.48 \times 10^{-16}\text{ m} \approx 2.95 \times 10^{-14}\text{ m}$$ ### Pattern Recognition Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z_1e)(Z_2e)/r_0$k(Z_1e)(Z_2e)/r_0$. Alpha particle has charge 2e$2e$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula E_n = fracE_0n^2$$E_n = \frac{E_0}{n^2}$$ L = fracnh2pi$$L = \frac{nh}{2\pi}$$ ### Core Logic From Bohr's theory, the energy of an electron in the n$n$-th orbit is inversely proportional to n^2$n^2$: E = -fracE_0n^2$$E = -\frac{E_0}{n^2}$$ Equating this to the given energy: -fracE_0n^2 = -0.04 E_0$$-\frac{E_0}{n^2} = -0.04 E_0$$ frac1n^2 = 0.04 = frac125$$\frac{1}{n^2} = 0.04 = \frac{1}{25}$$ n^2 = 25 implies n = 5$$n^2 = 25 \implies n = 5$$ ### Step 1: Final Conclusion Bohr's quantization condition for angular momentum: L = fracnh2pi$$L = \frac{nh}{2\pi}$$ Rearranging for the required term: frac2pi Lh = n$$\frac{2\pi L}{h} = n$$ Since n=5$n=5$, the value is 5$5$. ### Pattern Recognition The expression 2pi L / h$2\pi L / h$ is a direct request for the principal quantum number n$n$. E_n = E_1 / n^2$E_n = E_1 / n^2$ allows finding n$n$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula frac1lambda = R left( frac1n_1^2 - frac1n_2^2 right)$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ ### Core Logic For smallest wavelength of Lyman series (n_1 = 1, n_2 = infty$n_1 = 1, n_2 = \infty$): frac1lambda_L, textmin = R left(frac11^2 - 0right) implies R = frac191 mathrm~nm^-1$$\frac{1}{\lambda_{L, \text{min}}} = R \left(\frac{1}{1^2} - 0\right) \implies R = \frac{1}{91} \mathrm{~nm}^{-1}$$ For largest wavelength of Balmer series (n_1 = 2, n_2 = 3$n_1 = 2, n_2 = 3$): frac1lambda_B = R left(frac14 - frac19right) = frac191 times frac536 implies lambda_B = frac91 times 365 = 655.2 mathrm~nm$$\frac{1}{\lambda_{B}} = R \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{1}{91} \times \frac{5}{36} \implies \lambda_B = \frac{91 \times 36}{5} = 655.2 \mathrm{~nm}$$ For largest wavelength of Paschen series (n_1 = 3, n_2 = 4$n_1 = 3, n_2 = 4$): frac1lambda_P = R left(frac19 - frac116right) = frac191 times frac7144 implies lambda_P = frac91 times 1447 = 1872 mathrm~nm$$\frac{1}{\lambda_{P}} = R \left(\frac{1}{9} - \frac{1}{16}\right) = \frac{1}{91} \times \frac{7}{144} \implies \lambda_P = \frac{91 \times 144}{7} = 1872 \mathrm{~nm}$$ Difference in wavelengths: Delta lambda = lambda_P - lambda_B = 1872 - 655.2 = 1216.8 mathrm~nm approx 1217 mathrm~nm$$\Delta \lambda = \lambda_P - \lambda_B = 1872 - 655.2 = 1216.8 \mathrm{~nm} \approx 1217 \mathrm{~nm}$$ ### Step 1: Final Conclusion The wavelength difference is approximately 1217 mathrm~nm$1217 \mathrm{~nm}$. ### Pattern Recognition Spectral lines: Smallest wavelength rightarrow n_2 = infty$\rightarrow n_2 = \infty$. Largest wavelength rightarrow n_2 = n_1 + 1$\rightarrow n_2 = n_1 + 1$. Rydberg constant R = 1/lambda_L,textmin$R = 1/\lambda_{L,\text{min}}$. Substitute values into Balmer (2rightarrow 3$2\rightarrow 3$) and Paschen (3rightarrow 4$3\rightarrow 4$) equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula frac1lambda = Rleft(frac1n_1^2 - frac1n_2^2right)$$\frac{1}{\lambda} = R\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)$$ ### Core Logic Assess each statement using the Rydberg formula. Max wavelength corresponds to min energy (transition from n_1+1 to n_1$n_{1}+1 \to n_{1}$), and min wavelength corresponds to max energy (n_2 to infty$n_{2} \to \infty$). ### Step 1: Evaluate A and D (Lyman Series) For Lyman series, n_1 = 1$n_{1} = 1$. Maximum wavelength (n_2 = 2$n_{2} = 2$): frac1lambda_max = Rleft(1 - frac14right) = frac3R4 implies lambda_max = frac43R$$\frac{1}{\lambda_{\max}} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \implies \lambda_{\max} = \frac{4}{3R}$$ Statement A is correct. Minimum wavelength (n_2 = infty$n_{2} = \infty$): frac1lambda_min = Rleft(1 - 0right) = R implies lambda_min = frac1R$$\frac{1}{\lambda_{\min}} = R\left(1 - 0\right) = R \implies \lambda_{\min} = \frac{1}{R}$$ Statement D is incorrect. ### Step 2: Evaluate B (Balmer Series) Balmer series corresponds to transitions to n_1 = 2$n_{1} = 2$. These transitions primarily emit in the visible spectrum. Statement B is correct. ### Step 3: Evaluate C (Paschen Series) For Paschen series, n_1 = 3$n_{1} = 3$. Minimum wavelength (n_2 = infty$n_{2} = \infty$): frac1lambda_min = Rleft(frac19 - 0right) = fracR9 implies lambda_min = frac9R$$\frac{1}{\lambda_{\min}} = R\left(\frac{1}{9} - 0\right) = \frac{R}{9} \implies \lambda_{\min} = \frac{9}{R}$$ Statement C is correct. ### Step 4: Final Conclusion Statements A, B, and C are correct. ### Pattern Recognition Sees: "minimum/maximum wavelength" → Max wavelength = adjacent orbital drop (n+1$n+1$ to n$n$). Min wavelength = drop from infinity (infty$\infty$ to n$n$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms