hydrogen atom radius 5.29×10^(-11) m electron charge –1.60×10^(-19) C

hydrogen atom radius 5.29×10^(-11) m electron charge –1.60×10^(-19) C

 

 

Supposing that an electron in a hydrogen atom

is positioned at an average distance of 5.29×10^(-11) m

from the nucleus (so called Bohr radius of the atom),

determine the Coulomb force of attraction

between the electron and the proton

(that is, the hydrogen atom nucleus at the center of the atom).

The electron charge is –1.60×10^(-19) C,

the proton charge is +1.60×10^(-19) C.

 

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F = k |q1| |q2| / r^2

> k: Coulomb’s constant

( 참고 https://ywpop.tistory.com/8268 )

 

 

 

F = k q^2 / r^2

= (8.99×10^9 N•m^2/C^2) (1.60×10^(-19) C)^2 / (5.29×10^(-11) m)^2

= (8.99×10^9) (1.60×10^(-19))^2 / (5.29×10^(-11))^2

= 8.22×10^(-8) N

 

 

 

답: 8.22×10^(-8) N

 

 

 

 

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