Consider that an electron is revolving in an excited state of Hydrogen atom with velocity √(25.6) × 10⁵ m s⁻¹. The radius of the orbit is x × 10⁻⁹ m. The value of x is: [Take the mass of electron to the 9 × 10⁻³¹ kg, charge of electron = -1.6 × 10⁻¹⁹ C and (1)/(4πε₀) = 9 × 10⁹ N m²C⁻² ] (a) 1 (b) 4 (c) 3 (d) 2

Answer: (a). mv²/r = (ke²)/(r²) ⇒ r = (ke²)/(mv²) r = 9×10⁹×2.56×10⁻³⁸9×10⁻³¹×25.6×10¹⁰ = 2.304×10⁻²⁸2.304×10⁻¹⁹ r = 1×10⁻⁹ m So the answer is (a) 1.

Consider that an electron is revolving in an excited state of… | NEET 2026 (Re-neet)

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Bohr Model: Orbits, Speed & Angular Momentum

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^{5}\text{ m s}^{-1}$ . The radius of the orbit is $x \times 10^{-9}\text{ m}$ . The value of $x$ is:

[Take the mass of electron to the $9 \times 10^{-31}\text{ kg}$ , charge of electron $= -1.6 \times 10^{-19}\text{ C}$ and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ N m}^2\text{C}^{-2}$ ]

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Answer: (a)

$\frac{mv^2}{r} = \frac{ke^2}{r^2} \Rightarrow r = \frac{ke^2}{mv^2}$
$r = \frac{9\times10^9\times2.56\times10^{-38}}{9\times10^{-31}\times25.6\times10^{10}} = \frac{2.304\times10^{-28}}{2.304\times10^{-19}}$
$r = 1\times10^{-9}\text{ m}$
So the answer is (a) 1.

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