A electron of mass `9xx10^(-31)` kg is revolving in a stable circular orbit or radius `1.8xx10^(-12)` m in a hydrogen atom. If the electrostatic froce of attraction between the proton and electron is 8xx10^(-8)` N, the velosity of the electron is
A
`4xx10^(5) m//s`
B
`5xx10^(5)m//s`
C
`3xx10^(6)m//s`
D
`4xx10^(6)m//s`
Text Solution
Verified by Experts
The correct Answer is:
A
The electrostatic force of attraction (F ) . Between the proton and the electron provides the C.P force `(mv^(2))/(r )` for the motion of the electron . ` therefore F= (mv^(2))/(r ) ` ` therefore v^(2) = (F xxr ) /(m ) = (8xx10 ^(-8) xx1.8 xx10 ^(-12))/(9xx10^(-31))` `=1.6xx10^(11) = 16 xx10 ^(10)` ` therefore v= 4 xx10^(5) m//s`
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