På denna har jag haft rätt på A men får inte rätt på B
(a)
An electron and a 0.0500 kg bullet each have a velocity of magnitude 490 m/s, accurate to within 0.0100%. Within what lower limit could we determine the position of each object along the direction of the velocity? (Give the lower limit for the electron in mm and that for the bullet in m.)
for the electron
1.18
Correct: Your answer is correct.
mm
for the bullet
2.15e-32
Correct: Your answer is correct.
m
(b)
What If? Within what lower limit could we determine the position of each object along the direction of the velocity if the electron and the bullet were both relativistic, traveling at 0.300c measured with the same accuracy? (Give the lower limit for the electron in nm and that for the bullet in m.)
for the electron
0.598
Incorrect: Your answer is incorrect.
Again, you will need to use the uncertainty principle, but note now the velocity is high compared to the speed of light. So, you will need to use the relativistic definition of momentum. To find the uncertainty in velocity, treat the momentum and velocity uncertainties as differentials. This will require finding the derivative of relativistic momentum with respect to velocity. Also, be sure to express your answer in nanometers nm
for the bullet
1.09e-32
Incorrect: Your answer is incorrect.
Again, you will need to use the uncertainty principle, but note now the velocity is high compared to the speed of light. So, you will need to use the relativistic definition of momentum. To find the uncertainty in velocity, treat the momentum and velocity uncertainties as differentials. This will require finding the derivative of relativistic momentum with respect to velocity. m
Och detta är min andra uppgift
A photon having energy 0.790 MeV is scattered by a free electron initially at rest such that the scattering angle of the scattered electron is equal to that of the scattered photon as shown in the figure below.
An illustration shows the scattering of a photon by an electron. A photon labeled E0 is incoming horizontally from the left and hits the electron. The electron is scattered up and to the right at an angle 𝜃 above the horizontal, and a photon is scattered down and to the right at an angle 𝜃 below the horizontal.
(a) Determine the scattering angle of the photon and the electron.
photon
8.5
Incorrect: Your answer is incorrect.
What is the relationship between the momentum of the electron and the momentum of the scattered photon? °
electron
8.5
Incorrect: Your answer is incorrect.
You are correct that this angle is equal to 𝜃, but your result for 𝜃 is incorrect. °
(b) Determine the energy and momentum of the scattered photon.
energy
0.779
Incorrect: Your answer is incorrect.
If you know the wavelength of the photon you can find its energy. MeV
momentum
4.16e-22
Incorrect: Your answer is incorrect.
You appear to be calculating the momentum correctly using your incorrect result for the energy of the photon. kg · m/s
(c) Determine the kinetic energy and momentum of the scattered electron.
kinetic energy
0.011
Incorrect: Your answer is incorrect.
You have used energy conservation correctly, but your result for the energy of the scattered photon is incorrect. MeV
momentum
5.67e-24
Incorrect: Your answer is incorrect.
What is the relationship between the magnitudes of the momentum of the electron and the momentum of the photon? kg · m/s
(a)
An electron and a 0.0500 kg bullet each have a velocity of magnitude 490 m/s, accurate to within 0.0100%. Within what lower limit could we determine the position of each object along the direction of the velocity? (Give the lower limit for the electron in mm and that for the bullet in m.)
for the electron
1.18
Correct: Your answer is correct.
mm
for the bullet
2.15e-32
Correct: Your answer is correct.
m
(b)
What If? Within what lower limit could we determine the position of each object along the direction of the velocity if the electron and the bullet were both relativistic, traveling at 0.300c measured with the same accuracy? (Give the lower limit for the electron in nm and that for the bullet in m.)
for the electron
0.598
Incorrect: Your answer is incorrect.
Again, you will need to use the uncertainty principle, but note now the velocity is high compared to the speed of light. So, you will need to use the relativistic definition of momentum. To find the uncertainty in velocity, treat the momentum and velocity uncertainties as differentials. This will require finding the derivative of relativistic momentum with respect to velocity. Also, be sure to express your answer in nanometers nm
for the bullet
1.09e-32
Incorrect: Your answer is incorrect.
Again, you will need to use the uncertainty principle, but note now the velocity is high compared to the speed of light. So, you will need to use the relativistic definition of momentum. To find the uncertainty in velocity, treat the momentum and velocity uncertainties as differentials. This will require finding the derivative of relativistic momentum with respect to velocity. m
Och detta är min andra uppgift
A photon having energy 0.790 MeV is scattered by a free electron initially at rest such that the scattering angle of the scattered electron is equal to that of the scattered photon as shown in the figure below.
An illustration shows the scattering of a photon by an electron. A photon labeled E0 is incoming horizontally from the left and hits the electron. The electron is scattered up and to the right at an angle 𝜃 above the horizontal, and a photon is scattered down and to the right at an angle 𝜃 below the horizontal.
(a) Determine the scattering angle of the photon and the electron.
photon
8.5
Incorrect: Your answer is incorrect.
What is the relationship between the momentum of the electron and the momentum of the scattered photon? °
electron
8.5
Incorrect: Your answer is incorrect.
You are correct that this angle is equal to 𝜃, but your result for 𝜃 is incorrect. °
(b) Determine the energy and momentum of the scattered photon.
energy
0.779
Incorrect: Your answer is incorrect.
If you know the wavelength of the photon you can find its energy. MeV
momentum
4.16e-22
Incorrect: Your answer is incorrect.
You appear to be calculating the momentum correctly using your incorrect result for the energy of the photon. kg · m/s
(c) Determine the kinetic energy and momentum of the scattered electron.
kinetic energy
0.011
Incorrect: Your answer is incorrect.
You have used energy conservation correctly, but your result for the energy of the scattered photon is incorrect. MeV
momentum
5.67e-24
Incorrect: Your answer is incorrect.
What is the relationship between the magnitudes of the momentum of the electron and the momentum of the photon? kg · m/s
