Electron velocity from de Broglie wavelength and photon energy from wavelength

[Howlin]

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?

 

[rock.freak667]

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

$$\lambda = \frac{h}{mv}$$

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.