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Mashcka [7]
3 years ago
5

-In the Bohr model, as it is known today, the electron is imagined to move in a circular orbit about a stationary proton. The fo

rce 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
Physics
1 answer:
vampirchik [111]3 years ago
4 0

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

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