### Related Formula 2pi r_n = nlambda$2\pi r_n = n\lambda$ r_n = a_0 cdot n^2$r_n = a_0 \cdot n^2$ ### Core Logic According to Bohr's quantization postulate condition coupled with de-Broglie's hypothesis : 2pi r_n = nlambda$2\pi r_n = n\lambda$ For the second orbit (n = 2$n = 2$), the radius is: r_2 = a_0 cdot (2)^2 = 4a_0$r_2 = a_0 \cdot (2)^2 = 4a_0$ Substituting into the wave perimeter formula : 2pi (4a_0) = nlambda$2\pi (4a_0) = n\lambda$ lambda = frac8pi a_0n$\lambda = \frac{8\pi a_0}{n}$ This strictly matches option (2). ### Pattern Recognition The circumference of an electron's orbit must encompass an exact integral count of complete standing de-Broglie wave wavelengths. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom