The distance between electron and proton in a hydrogen atom is 0.53 A. Find the speed and frequency of revolution of electron round the proton. (Take. ` m_e= 9.1 xx 10 ^(-31) kg) `

Explanation

Here ` F_e= (mv^(2))/(r) `
` rArr (kq_1q_2)/(r^(2)) = (mv^(2))/(r) rArr v=sqrt((kq^(2))/(mr)) = q sqrt((k)/(mr)) `
` " " = 1.6 xx 10 ^(-19) xx sqrt((9xx 10 ^(9))/(0.53 xx 10 ^(-10) xx 9.1 xx 10 ^(_31)))`
` = 2.19 xx 10 ^(6) m//s`
As, v = r . omega = r (2pin)" " (because omega =2pi n )`
` therefore " " n= (v)/(2pir ) = (2.19xx 10^(6))/(2 xx 3.14 xx 0.53 xx 10 ^(-10)) `
` = 0.65 xx 10^(16) =6.5 xx10 ^(15) Hz`