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Physics · PYQ NEET UG 2026
हिन्दीConsider that an electron is revolving in an excited state of Hydrogen atom with velocity ( \sqrt{25.6} \times 10^5 \text{ ms}^{-1} ). The radius of the orbit is ( x \times 10^{-9} \text{ m} ). The value of ( x ) is : [Take the mass of electron to be ( 9 \times 10^{-31} \text{ kg} ), charge of electron ( = -1.6 \times 10^{-19} \text{ C} ) and ( \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2} )]
Correct answer · Option 3
2
Explanation
Using Bohr's quantization condition and Coulomb force balancing centripetal force, the radius is calculated as ( r = \frac{1}{4\pi \epsilon_0} \frac{e^2}{m v^2} ), yielding ( x = 2 ).
