Question

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.84

Answer 1

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

Answer 2

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

Answer 3

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