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Question

The radius of the first orbit in a hydrogen atom is \( 5.3 \times 10^{-11} \, \text{m} \). What is the
radius of the fifth orbit?

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Explanation

**Excitation** energy required to go from n=1 to n=3 is 12.09 eV, from ground to n=∞ ionization 13.6 eV, state n=∞ means ionized, electron free with zero energy, highest level reached by electron beam energy determines which levels can be excited, e.g., 11 eV beam from ground can reach n=2 (10.2 eV) but not n=3 (12.09 eV), so max n=2. r_n = n² r₁ . For n = 5 : r₅ = 5² × 5.3 × 10⁻¹¹ = 25 × 5.3 × 10⁻¹¹ = 1.325 × 10⁻⁹ m . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀

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