HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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The Electron in the Hydrogen Atom is Moving with a Speed of 2.3106 M/S in an Orbit of Radius 0.53 å. Calculate the Period of Revolution of the Electron. (π = 3.142) - HSC Science (Electronics) 12th Board Exam - Physics

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Question

The electron in the hydrogen atom is moving with a speed of 2.3x106 m/s in an orbit of radius 0.53 Å. Calculate the period of revolution of the electron. (Π = 3.142)

Solution

`v = 2.3xx10^6 m`

`r = 0.53Å = 0.53 xx 10^(-10)m`

`v = romega = r xx (2pi)/T`

`T = (2pir)/v = (2xx3.14xx0.53xx10^
(-10))/(2.3 xx10^6)`

`:. T = 1.44 xx 10^(-16)` sec

`f = 1/T = 6.9 xx 10^(15)` Hz

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APPEARS IN

 2017-2018 (March) (with solutions)
Question 6.5 | 2.00 marks
Solution The Electron in the Hydrogen Atom is Moving with a Speed of 2.3106 M/S in an Orbit of Radius 0.53 å. Calculate the Period of Revolution of the Electron. (π = 3.142) Concept: Magnetic Dipole Moment of a Revolving Electron.
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