Orbital Frequency of an electron in a hydrogen atom

  • #1

Homework Statement



In a classical model of the hydrogen atom, the electron moves around the proton in a circular orbit of radius 0.053 nm.

What is the electron's orbital frequency?

What is the effective current of the electron?


Homework Equations



Freq * Wavelength = Speed of light
(V*lambda = c)

Lambda = (plancks constant)/momentum

momentum = (mass of electron)*(Velocity)

F = qE = m(v^2)/R

current....I = q*ie....(charge of electron)*(electron current)

number of electrons...Ne = ie*delta_t....(electron current)*(period I presume)

Ne = 1

The Attempt at a Solution



Mass of electron = 9.1094*10^-31 kg
qe- = -1.6*10^-19 coulombs
radius = 0.053*10^-9 m

...utlizing this information I found the velocity from the Force equation, deriving:
sqrt(K(q^2)/(r*me-) = v

Then plugged in velocity into the equation: lambda = h/(mv)

Plugged lambda into: V = c/lambda

....finding my frequency to be about 900*10^15 Hz....but this was wrong. Help would be extremely appreciated!!!
 

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
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Freq * Wavelength = Speed of light
(V*lambda = c)
This is the equation for the frequency of a light wave, which is not really relevant here.

The question is asking you for the orbital frequency of the electron i.e. how often does a full cycle (orbit) repeat?

Well, the orbital frequency is just the reciprocal of the orbital period. How do you figure out the orbital period? It's as simple as remembering that distance = speed*time. What distance is covered by the electron in one orbit?
 
  • #3
ohhh.......well, I overthunk it. Lol! So, take the velocity I found, and divide it into the circumference: (2*pi*r)/v = seconds....in other words the period. THANKS!! It worked!
 

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