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

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SUMMARY

This discussion focuses on calculating the velocity of an electron in an electron microscope and the energy of photons emitted by sodium lights. The correct electron velocity, derived using De Broglie's wavelength equation, is 5.95E7 m/s, contrasting with the incorrect assumption of 3E8 m/s. For photon energy, the calculation using the formula E=hf yields an energy of approximately 4.3E-7 eV for photons with a wavelength of 580 nm, correcting the initial miscalculation of 2.107 eV.

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  • Understanding of De Broglie's wavelength equation
  • Familiarity with the Planck-Einstein relation (E=hf)
  • Knowledge of wave properties of particles
  • Basic concepts of electron microscopy
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Howlin
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Homework Statement


A) The wavelength 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/wavelength
w=2pi*f


The Attempt at a Solution


A) k=2pi/0.0122nm = 5.15x10^+11
w=2pi(c/wavelength) = (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/wavelength) = (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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