Answer:
a) 4.04*10^-12m
b) 0.0209nm
c) 0.253MeV
Explanation:
The formula for Compton's scattering is given by:
![\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)](https://tex.z-dn.net/?f=%5CDelta%20%5Clambda%3D%5Clambda_f-%5Clambda_i%3D%5Cfrac%7Bh%7D%7Bm_oc%7D%281-cos%5Ctheta%29)
where h is the Planck's constant, m is the mass of the electron and c is the speed of light.
a) by replacing in the formula you obtain the Compton shift:
![\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m](https://tex.z-dn.net/?f=%5CDelta%20%5Clambda%3D%5Cfrac%7B6.62%2A10%5E%7B-34%7DJs%7D%7B%289.1%2A10%5E%7B-31%7Dkg%29%283%2A10%5E8m%2Fs%29%7D%281-cos132%5C%C2%B0%29%3D4.04%2A10%5E%7B-12%7Dm)
b) The change in photon energy is given by:
![\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm](https://tex.z-dn.net/?f=%5CDelta%20E%3DE_f-E_i%3Dh%5Cfrac%7Bc%7D%7B%5Clambda_f%7D-h%5Cfrac%7Bc%7D%7B%5Clambda_i%7D%3Dhc%28%5Cfrac%7B1%7D%7B%5Clambda_f%7D-%5Cfrac%7B1%7D%7B%5Clambda_i%7D%29%5C%5C%5C%5C%5Clambda_f%3D4.04%2A10%5E%7B-12%7Dm%20%2B%5Clambda_i%3D4.04%2A10%5E%7B-12%7Dm%2B%280.0169%2A10%5E%7B-9%7Dm%29%3D2.09%2A10%5E%7B-11%7Dm%3D0.0209nm)
c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.
![P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bh%7D%7B%5Clambda_e%7D%3D%5Cfrac%7B6.62%2A10%5E%7B-34%7DJs%7D%7B2.43%2A10%5E%7B-12%7Dm%7D%3D2.72%2A10%5E%7B-22%7Dkgm%5C%5C)
![E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV](https://tex.z-dn.net/?f=E_e%3D%5Cfrac%7Bp%5E2%7D%7B2m_e%7D%3D%5Cfrac%7B%282.72%2A10%5E%7B-22%7Dkgm%29%5E2%7D%7B2%289.1%2A10%5E%7B-31%7Dkg%29%7D%3D4.06%2A10%5E%7B-14%7DJ%5C%5C%5C%5C1J%3D6.242%2A10%5E%7B18%7DeV%5C%5C%5C%5CE_e%3D4.06%2A10%5E%7B-14%7D%286.242%2A10%5E%7B18%7DeV%29%3D0.253MeV)
Answer:
Explanation:
you can seperate mixtures by filtration, evaporation, distillation and chromatography.
Answer:
(A) more rapidly than
Explanation:
With higher temperatures, object's molecules (and atoms) have higher kinetic energy which is due to faster "jiggling" (vibrations). On a hot day these vibrations in the material the sidewalk is made of are more rapid than on a cold day, just as their temperatures differ.
-- Accelerating at the rate of 8 m/s², Andy's speed
after 30 seconds is
(8 m/s²) x (30.0 s) = 240 m/s .
-- His average speed during that time is
(1/2) (0 + 240 m/s) = 120 m/s .
-- In 30 sec at an average speed of 120 m/s,
Andy will travel a distance of
(120 m/s) x (30 sec) = 3,600 m
= 3.6 km .
"But how ? ! ?", you ask.
How in the world can Andy leave a stop light and then
cover 3.6 km = 2.24 miles in the next 30 seconds ?
The answer is: His acceleration of 8 m/s², or about 0.82 G
is what does it for him.
At that rate of acceleration ...
-- Andy achieves "Zero to 60 mph" in 3.35 seconds,
and then he keeps accelerating.
-- He hits 100 mph in 5.59 seconds after jumping the light ...
and then he keeps accelerating.
-- He hits 200 mph in 11.2 seconds after jumping the light ...
and then he keeps accelerating.
-- After accelerating at 8 m/s² for 30 seconds, Andy and his
car are moving at 537 miles per hour !
We really don't know whether he keeps accelerating,
but we kind of doubt it.
A couple of observations in conclusion:
-- We can't actually calculate his displacement with the information given.
Displacement is the distance and direction between the starting- and
ending-points, and we're not told whether Andy maintains a straight line
during this tense period, or is all over the road, adding great distance
but not a lot of displacement.
-- It's also likely that sometime during this performance, he is pulled
over to the side by an alert cop in a traffic-control helicopter, and
never actually succeeds in accomplishing the given description.