Answers
The speed of the electron in the orbit is
v = nh/ 2 pi m r = 1( 6.63 * 10 ^-34 Js)/ 2 pi ( 9.1 * 10 ^-31 kg) (5.84 * 10 ^-11 m)
=1.98* 10 ^ 10 m/s
.F = qE = -m(v^2/r)
(note acc. is minus because its towards the centre)
so qE = -m(v^2/r) here q =e charge of proton =1..6*10^-9 m =mass of electron =9.109*10^-31kg and k =9*10^9 constant
E =-k e /r^2
ke^2/(r^2) =mv^2/r
put the value
(9*10^9)(1.6*10^-19)^2 / (5.84*10^-11)^2 =(9.109*10^-31) v^2 /(5.84* 10^-11)
after solving
v =6.58*10^6 m/s answer
apply Kinetic energy KE = chnage in pE
i.e 0.5 mv^2 = Kq1q2/r
where m is mass of electron
v is speed = ?
K is consatnt = 9e9
q1 q2 are the magnitudes of charges = 1.6 e-19 C each
and r is distancr = 5.84 e-11 m
so
speed v^2 = 2Kq1q2/mr
v^2 = 2* 9e9 * 1.6 e-19*1.6 e-19/(9.11e-31 * 5.84 e-11)
v^2 = 8.166 *10^12
v = 2.048 *10^6 m/.s ---------------<<<<<<<<answer
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