In a hydrogen atom, an electron of mass `9 . 1 xx 10^(-31)` kg revolves about a proton in circular orbit of radius 0 . 53 `Å` . The radial acceleration and angular velocity of electron are respectively .

Options

A. `9 xx 10^(22) ms^(-2), 4 . 1 xx 10^(16) s^(-1)`
B. `4 . 1 xx 10^(16 ms^(-2) , 9 xx 10^(22) s^(-1)`
C. `9 xx 10^(16) ms^(-2) , 4.1 xx 10^(22) s^(-1)`
D. `4 . 1 xx 10^(22) ms^(-2) , 9 xx 10^(16) s^(-1)`

Answer

A

Explanation

Given,mass of electron, `m_(e) = 9 . 1xx 10^(-31) kg` and radius of circular orbit, r ` = 0 . 53 xx 10^(-10) m `
`:.` An electron revolves about a proton in circular orbit . So , centripetel force- electrostatic force
`(m_(e)v^(2))/(r) = (kq_(1)q^(2))/(r^(2))`
`m_(e) v^(2)r = kq_(1)q_(2)`
`:. v = r omega`
`:. m_(e) (r omega)^(2) = kq_(1) q_(2)`
`rArr omega^(2) = (kq^(2))/(m_(e)r^(3)) rArr omega = sqrt((kq^(2))/(m_(e)r^(3)))`
`:. ` angular velocity of electron,
`omega = sqrt((9 xx 10^(9) xx (1.6xx 10 ^(-19))^(2))/(9.1 xx 10^(-31) xx (0.53 xx 10^(-10))^(-3)))`
`omega = 4.1 xx 10^(16) s^(-1)`
Hence, the radial acceleration of electron = `m r omeha ^(2)`
`= 0 . 53 xx 10^(-10)xx (4.1 xx 10^(16))^(2) = 9 xx 10^(22) ms^(-2)`