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\(a^2 = b_0\) \(r_n = \left ( \frac{n^2}{Z} \right ) a_0\), where \(a_0\) = 0.529 \(A°\) \( K.E. = \left( \frac{Z^2}{n^2} \right )2.18 \times 10^{-18} J.atom^{-1}\)
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For JEE (Mains and Advanced) Physics 12 – Syllabus for JEE Main Physics 12 – Syllabus for JEE Advanced Scope of these Syllabi