An electron in the ground state of the hydrogen atom has the orbital radius of 5.3 × 10⁻¹¹ m while that for the electron in third excited state is 8.48 × 10⁻¹⁰ m. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is (a) 4 (b) 3 (c) 9 (d) 16

Answer: (a). r₄/r₁ = 8.48×10⁻¹⁰5.3×10⁻¹¹ = 16 = n², so the excited state is n = 4. λ = (2π rn)/(n) ∝ n λgroundλexcited = 1/4, i.e. λexcited = 4λground. The key takes the magnitude 4, option (a).

An electron in the ground state of the hydrogen atom has the orbital… | JEE Main 2025 (22 Jan, Shift 1)

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Matter Waves: de Broglie Wavelength

An electron in the ground state of the hydrogen atom has the orbital radius of $5.3 \times 10^{-11}\text{ m}$ while that for the electron in third excited state is $8.48 \times 10^{-10}\text{ m}$ . The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is

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Answer: (a)

$r_4/r_1 = \dfrac{8.48\times10^{-10}}{5.3\times10^{-11}} = 16 = n^2$ , so the excited state is $n = 4$ .
$\lambda = \dfrac{2\pi r_n}{n} \propto n$
$\dfrac{\lambda_{ground}}{\lambda_{excited}} = \dfrac{1}{4}$ , i.e. $\lambda_{excited} = 4\lambda_{ground}$ .
The key takes the magnitude 4, option (a).

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