The presence of oxygen atoms on the surface results in auger electrons with a kinetic energy of approximately 505 eV.
What is the de Broglie wavelength of this electron?
[KE= (1/2)mv^2; 1 electron volt (eV) = 1.602*10^-19 J]
p= sqrt (KE*m*2)
= sqrt(505*1.602E-19*9.11E-31*2)
then
lambda= plancks constant/p
What is the wavelength (in meters) of a proton ({\rm{mass}} = 1.673 \times 10^{ - 24} {\rm g}) that has been accelerated to 28% of the speed of light?
To find the de Broglie wavelength of an electron with a kinetic energy of 505 eV, we can use the de Broglie wavelength formula:
λ = h / p
where λ is the de Broglie wavelength, h is the Planck's constant (approximately 6.626 * 10^-34 J·s), and p is the momentum of the electron.
The momentum of an electron can be calculated using the kinetic energy and the mass of the electron:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass of the electron (approximately 9.10938356 × 10^-31 kg), and v is the velocity of the electron.
First, let's convert the kinetic energy from eV to joules:
1 eV = 1.602 * 10^-19 J
So, the kinetic energy (KE) in joules is:
KE = 505 eV * (1.602 * 10^-19 J/eV) = 8.09 * 10^-17 J
Now, we can calculate the velocity of the electron by rearranging the kinetic energy formula:
KE = (1/2)mv^2
v = √(2KE / m)
v = √(2 * 8.09 * 10^-17 J / (9.10938356 × 10^-31 kg))
v ≈ 1.380 * 10^6 m/s
Next, we can calculate the momentum using the formula:
p = mv
p = (9.10938356 × 10^-31 kg) * (1.380 * 10^6 m/s)
p ≈ 1.258 * 10^-24 kg·m/s
Finally, we can use the de Broglie wavelength formula to find λ:
λ = h / p
λ = (6.626 * 10^-34 J·s) / (1.258 * 10^-24 kg·m/s)
λ ≈ 5.27 * 10^-10 meters
Therefore, the de Broglie wavelength of the electron with a kinetic energy of approximately 505 eV is approximately 5.27 * 10^-10 meters.