Experimentally it is found that 12.8 eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is 9/x × 10⁻¹⁰ m. The value of the x is __________. (1 eV = 1.6 × 10⁻¹⁹ J, (1)/(4πε₀) = 9 × 10⁹ Nm²/C² and electronic charge = 1.6 × 10⁻¹⁹ C) Numerical answer type. Work it out, then open the answer.

Answer: 16. Ionisation energy = (ke²)/(2r) ⇒ r = (ke²)/(2E) r = 9×10⁹×(1.6×10⁻¹⁹)²2×12.8×1.6×10⁻¹⁹ = 14.4×10⁻¹⁰25.6 r = 0.5625×10⁻¹⁰ = 9/16×10⁻¹⁰ m So x = 16.

Experimentally it is found that 12.8 eV energy is required to… | JEE Main 2023 (6 Apr, Shift 2)

Q71JEE Main 2023 · 6 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum

Experimentally it is found that $12.8\text{ eV}$ energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is $\frac{9}{x} \times 10^{-10}\text{ m}$ . The value of the $x$ is __________.

$\Bigg(1\text{ eV} = 1.6 \times 10^{-19}\text{ J},\ \frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ Nm}^2/\text{C}^2$ and electronic charge $= 1.6 \times 10^{-19}\text{ C}\Bigg)$

Numerical answer type. Work it out, then open the answer.

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Answer: 16

Ionisation energy $= \frac{ke^2}{2r} \Rightarrow r = \frac{ke^2}{2E}$
$r = \frac{9\times10^9\times(1.6\times10^{-19})^2}{2\times12.8\times1.6\times10^{-19}} = \frac{14.4\times10^{-10}}{25.6}$
$r = 0.5625\times10^{-10} = \frac{9}{16}\times10^{-10}\text{ m}$
So $x = 16$ .

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