Maple TA 25_2

Here's a head start on 25_2

1. A photon has a wavelength of 655.2 nm. What is the photon momentum in eV/c (sheet 14). Indicate with a positive (negative) sign whether the momentum unit quoted is (not) an SI unit.

= (6.64E-34)*(3E8)
=(1.98E-25)
THEN
=(1.98E-25)/(655.2E-9)
=(3.02198E-19)
THEN
=(3.02198E-19)/(1.6E-19)
=1.88874

Reasoning: First step is to get the value for value for (hc) then you divide by the wavelength to get the value of E, and lastly divide by 1.6E-19 to get the value in terms of eV instead of Joules.

BTW the answer for # 1 is NEGATIVE

Still trying to figure out # 2 and # 3

4. An atom makes a transition from its current state into an energy state which is 2.067 eV higher. Indicate with a negative (positive) sign whether a photon is emitted (absorbed) by the atom and calculate the wavelength of the photon (sheet 32).

=(6.64E-34)*(3E8)
=(1.992E-25)
THEN
=(1.992E-25)/(1.6E-19)
=(1.245E-6)
THEN
=(1.245E-6)/(2.067)
=(6.02322E-7)(POSITIVE)

2. A 17.37 KeV photon scatters off an electron (Compton Effect) and has a wavelength of 0.099 nm after the collision. What is the kinetic energy of the recoil electron in KeV (sheet 15-18) ? Indicate with a positive (negative) sign whether the electron kinetic energy is of the same order of magnitude as (considerably smaller than) the electron rest energy.

=((6.64E-34)*(3E8))/(.099E-9)
=(2.01212E-15)
THEN
=(2.01212E-15)/(1.6E-19)
=(12575.8eV)---->12.5758 KeV (in terms of KeV is what the question is asking for)
THEN
=(17.37) - (12.5758)
= (4.7942)(NEGATIVE)

Reasoning: You need to first solve to get "E", and then you divide that answer by 1.6E-19 to convert into eV, and then divide by 1000 to convert from eV to KeV. Then to find the kinetic energy of the recall electron, you subtract the two.

I'm having some trouble with # 3, if anyone can help with that one, please.

Still having no luck with # 3, if anyone can help.

Any ideas on number 3 I cant figure it out either?

first divide theta by 2 and convert the d into meters
lambda = (2dsin(theta))/n

second p = h/lambda from above

third KE= (p from above)^2/(2*9.1*10^-31)

forth convert to eV KE from above /(1.6*10^-19)

What is n in the first formula provided?


An electron beam scattered off a crystal at a scattering angle of 26.99 0 produces a second order diffraction maximum. What is the kinetic energy (nonrelativistically) of the electrons in eV if the crystal plane spacing is 0.2637 nm (sheet 22,27) ? Indicate with a positive (negative) sign whether here the particle (wave) property of the electron becomes evident.

what i did so far was 2*.2637E-9= 5.274E10*sin(13.495)/n, but what is n?

can someone please help with number three i dont understand how to solve it?

I am having trouble with that one as well if someone could post a more detailed description of how to solve it I would appreciate it. Thanks.

sck105....what do you mean by more details? i thought the details above by userx was very detailed lol.

Here's a full description on question 3.
An electron beam scattered off a crystal at a scattering angle of 27.9 0 produces a second order diffraction maximum. What is the kinetic energy (nonrelativistically) of the electrons in eV if the crystal plane spacing is 0.2926 nm (sheet 22,27) ? Indicate with a positive (negative) sign whether here the particle (wave) property of the electron becomes evident.
=(27.9)/2
=13.95
THEN
=(2)*(.2926E-9)*sin(13.95)
=(4.70665E-24)/2
=7.05386E-11
THEN
=(6.64E-34)/(7.05386E-11)
=9.41329E-29
THEN
=(9.41329E-29)^2
=8.861E-47
THEN
=(8.861E-47)/(2*9.1E-31)
=4.86868E-17
THEN
=(4.86868E-17)/(1.6E-19)
=304.293 (NEGATIVE)