in an electron microscope, a tungstun cathode with work function 4.5 eV is heated to release electrons that are then initially at rest just outside the cathode. The electrons are accelerated by a potential difference to create a beam of electrons with a de Broglie wavelength of 0.038 nm. Assume nonrelativistic apply to the motion of the electrons. calculate the momentum of an electron in the beam, in kg x m/s. calcualte the kinetic energy of an electron in the beam, in joules. calculate the accelerating voltage. suppose the light, instead of heat, is used to release the electrons from the cathode. What minimum frequency of light is needed to accomplish this?
There are many parts to this question, and several different physics equations are needed to do the parts.
I suggest you review how the De Broglie wavelength is related to momentum and Plank's constant, h. That will tell you the momentum, m V. Use the momentum and the electron mass m to get the kinetic energy, KE = (1/2) m V^2. The accelerating voltage V is given by KE = e *V, where e is the electron charge.
Review the photoelectric effect for the last part. Someone will be happy to critique your work.
To calculate the momentum of an electron in the beam, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength:
p = h / λ
Where:
p represents the momentum of the electron,
h is the Planck's constant (h ≈ 6.626 x 10^-34 J·s), and
λ is the de Broglie wavelength of the electron.
Given that the de Broglie wavelength of the electron is 0.038 nm (or 3.8 x 10^-11 m), we can substitute these values into the formula:
p = (6.626 x 10^-34 J·s) / (3.8 x 10^-11 m) ≈ 1.744 x 10^-23 kg·m/s
Therefore, the momentum of an electron in the beam is approximately 1.744 x 10^-23 kg·m/s.
To calculate the kinetic energy of an electron in the beam, we can use the kinetic energy formula:
K.E. = (1/2) mv^2
Where:
K.E. represents the kinetic energy of the electron,
m is the mass of the electron (m ≈ 9.109 x 10^-31 kg), and
v is the velocity of the electron.
Since the electrons are non-relativistic and initially at rest, the velocity of the electron can be approximated using the relation between velocity, momentum, and mass:
v = p / m
Substituting the values for p and m into the equation:
v = (1.744 x 10^-23 kg·m/s) / (9.109 x 10^-31 kg) ≈ 1.913 x 10^7 m/s
Now we can calculate the kinetic energy using the kinetic energy formula:
K.E. = (1/2) (9.109 x 10^-31 kg) (1.913 x 10^7 m/s)^2 ≈ 1.742 x 10^-17 J
Therefore, the kinetic energy of an electron in the beam is approximately 1.742 x 10^-17 J.
To calculate the accelerating voltage, we can use the equation that relates the kinetic energy of a charged particle to the potential difference:
K.E. = qV
Where:
K.E. is the kinetic energy of the electron,
q is the charge of the electron (q ≈ 1.602 x 10^-19 C), and
V is the accelerating voltage.
Rearranging the equation to solve for V:
V = K.E. / q
Substituting the values for K.E. and q into the equation:
V = (1.742 x 10^-17 J) / (1.602 x 10^-19 C) ≈ 108.729 V
Therefore, the accelerating voltage is approximately 108.729 V.
Now, let's consider the case where light is used to release electrons from the cathode. To determine the minimum frequency of light required, we can use Einstein's photoelectric equation:
E = hf - φ
Where:
E represents the energy of the incident photon,
h is Planck's constant (h ≈ 6.626 x 10^-34 J·s),
f is the frequency of light,
φ is the work function of the cathode material (φ ≈ 4.5 eV), and
eV is the electron volt (eV ≈ 1.602 x 10^-19 J).
Since we want to find the minimum frequency of light, we can rearrange the formula to solve for f:
f = (E + φ) / h
Substituting the values for E (since E = hf) and φ into the equation:
f = (hf + φ) / h = (E + φ) / h
Converting the work function from eV to joules:
φ = (4.5 eV) (1.602 x 10^-19 J/eV) ≈ 7.21 x 10^-19 J
Substituting the values for h and φ into the equation:
f = (6.626 x 10^-34 J·s) / [(1.742 x 10^-17 J) + (7.21 x 10^-19 J)] ≈ 2.9 x 10^15 Hz
Therefore, the minimum frequency of light required to release the electrons from the cathode is approximately 2.9 x 10^15 Hz.