How to Calculate Electron Velocity and Photon Energy in Physics Homework?

AI Thread Summary
To calculate electron velocity and photon energy, the discussion highlights the use of De Broglie's wavelength equation for electrons, emphasizing that the velocity should not be assumed to be the speed of light. For the electron in the microscope, the correct velocity is calculated to be approximately 5.95E7 m/s, not 3E8 m/s as initially stated. For the sodium light photons, the energy calculation using E=hf yields a result of 4.3E-7 eV, correcting the initial attempt. The importance of using appropriate formulas and constants for accurate results is stressed throughout the discussion. Accurate calculations are essential for understanding concepts in physics homework.
Howlin
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Homework Statement


A) The wavelenght of an electron in an electron microscope is 0.0122nm.What is the electron velocity?
B) Sodium lights emit light of yellow colour, with photons of wavelength 580nm. What is the energy of these photons?

Homework Equations


v=w/k
k=2pi/wavelenght
w=2pi*f


The Attempt at a Solution


A) k=2pi/0.0122nm = 5.15x10^+11
w=2pi(c/wavelenght) = (2pi*3E8)/0.0122nm
w=1.545E19
v=w/k=3E8 m/s

B)E=hf = hc/lambda
e=(4.136E-15)(3E8)/589E-9
e=2.107eV

The answer to A should be 5.95E7 m/s and the answer to B should be 4.3E-7eV. Am i right in my answers or are the teachers answers wrong?
 
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Howlin said:

The Attempt at a Solution


A) k=2pi/0.0122nm = 5.15x10^+11
w=2pi(c/wavelenght) = (2pi*3E8)/0.0122nm
w=1.545E19
v=w/k=3E8 m/s

For this, I think you'd want to use De Broglie's wavelength equation

\lambda = \frac{h}{mv}

Howlin said:
B)E=hf = hc/lambda
e=(4.136E-15)(3E8)/589E-9
e=2.107eV

You'd need to get the velocity using the same means as before, except the electrons energy will be kinetic.


Your main error lies in the fact that you are using the velocity of the electrons to be the speed of light ,3(108) m/s. This is incorrect for electrons.
 
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