Under appropriate conditions, molybdenum emits X rays that have a characteristis wavelength of 0.711 Angstrom.
These X rays are used in diffraction experiments to dtermine the structures of molecules. How fast would an electron have to be moving in order to have the same wavelength as these X rays?
The answer is V= 1.02x10^7 m/s
My question: At first I tried converting Angstroms to Meters and then using the formula V= C/Lambda, in which C is 3x10^7m/s and Lambda would be the Angstroms I coverted, however the m's x-out and then I would only have s....so I need some help!
Use the deBroglie hypothesis
wavelength= h/mv
solve for v.
By the way, the speed of light in free space is 3*108, not 107 m/s
To calculate the speed of an electron required to have the same wavelength as the X-rays emitted by molybdenum, you can use the equation:
λ = h / mV
where:
λ is the wavelength of the X-rays in meters,
h is the Planck's constant (approximately 6.626 x 10^-34 J s),
m is the mass of the electron (approximately 9.11 x 10^-31 kg), and
V is the velocity of the electron.
First, convert the given wavelength from Angstroms to meters:
λ = 0.711 Angstrom * (1 x 10^-10 m / 1 Angstrom)
= 7.11 x 10^-11 m
Now, we can rearrange the equation to solve for V:
V = h / (m * λ)
V = (6.626 x 10^-34 J s) / [(9.11 x 10^-31 kg) * (7.11 x 10^-11 m)]
= 1.02 x 10^7 m/s
Therefore, the speed of the electron would have to be approximately 1.02 x 10^7 m/s to have the same wavelength as the X-rays emitted by molybdenum.