Question

-In the Bohr model, as it is known today, the electron is imagined to move in a circular orbit about a stationary proton. The force responsible for the electron's circular motion is the electric force of attraction between the electron and the proton. If the speed of the electron were 2.4×105 m/s , what would be the corresponding orbital radius?

-The answer I got was 4x10^-9 , but it is wrong.

(9.0x109) (1.6x10-19 C)2 == 2.304x10-28 Then I divided that by (9.11x10-31 kg)( 2.4 x 105)2 =5.24736 x 10-20

=== 2.304x10-28 / 5.24736 x 10-20 = .000000004

I do not know what I am doing wrong. Please help

Answer 1

`4.3859007196* 10^(-9)\ m`

Step-by-step explanation:

k = Coulomb constant =
`8.99* 10^(9)\ Nm^2/C^2`

v = Velocity of electron =
`2.4* 10^5\ m/s`

q = Charge of electron =
`1.6* 10^(-19)\ C`

m = Mass of electron =
`9.11* 10^(-31)\ kg`

r = Radius

The electrical and centripetal force will balance each other

`(kq^2)/(r^2)=(mv^2)/(r)\\\Rightarrow r=(kq^2)/(mv^2)\\\Rightarrow r=(8.99* 10^9* (1.6* 10^(-19))^2)/(9.11* 10^(-31)* (2.4* 10^5)^2)\\\Rightarrow r=4.3859007196* 10^(-9)\ m`

The radius of the orbital is
`4.3859007196* 10^(-9)\ m`