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

Answer:

$4.3859007196\times 10^{-9}\ m$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

v = Velocity of electron = $2.4\times 10^5\ m/s$

q = Charge of electron = $1.6\times 10^{-19}\ C$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

r = Radius

The electrical and centripetal force will balance each other

$\dfrac{kq^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{kq^2}{mv^2}\\\Rightarrow r=\dfrac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times (2.4\times 10^5)^2}\\\Rightarrow r=4.3859007196\times 10^{-9}\ m$

The radius of the orbital is $4.3859007196\times 10^{-9}\ m$