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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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