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Electron Position With The Atomic Time [Bohr Model] [Fx-3650Pii]


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#1 frankmar98

frankmar98

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Posted 24 April 2016 - 06:53 PM

ClrMemory:
?=>C:
?=>D:
5.29E-11*C^2=>B:
(D*2.419E-17*C*6.63E-34)/(2*pi*9.11E-31*B^2)=>A:
B*cos(A)=>X:
B*sin(A)=>Y:
X(output triangle)
Y(output triangle)

All physic constants and magnitudes are in the International System of Units, except when you input the time, that is in the atomic unit of time (but I convert it to seconds to be coherent).

 

Angles are in radians, if the calculator is in degrees, it doesn´t work.

 

The constansts and formulas used are:

 

S=r*angle(rad)

 

v=S/t

 

m*v*r=n*(h/2pi)

 

r_c = (r * cos a, r * sin a)

 

Bohr´s first atomic radius = 5.29E-11 m

 

Mass of the electron = 9.11E-31 kg

 

Planck´s constant = 6.63E-34 J*s

 

1 second = 2.419E+17 Atomic Time Units

 

The results are displayed in meters, and the (0,0) point is the nucleous of the atom.

 

 

Variables:

 

C: Atomic Bohr´s level (of the electron that we want to study) (adimensional)

D: Time (ATU)

B: Electron´s orbit radious (meters)

A: Angle of the electron (considering that when time is 0, angle is 0) (rad)

X and Y: Electron positon (meters)

 

 

 

Regards!

 

P.D: I´ve edited it to complete it


Edited by frankmar98, 25 April 2016 - 01:58 PM.





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