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Physics 250
Practice Exam II
- In a Compton scattering event, the scattered photon has an energy of 120 keV and the recoiling electron has an energy of 40 keV. Find (a) the wavelength of the incident photon, (b) the angle at which the photon is scattered, and (c) the recoil angle of the electron.
(a)
(b)
(c)
E γ 2 =120 keV
Te = 40 keV
E γ 2
E (^) e T (^) e
E γ 1
φ
θ
θ φ
E γ 1 + m (^) e c^2 = E γ 2 + Te + m (^) e c^2 E γ 1 = 120 keV +40 keV =160 keV
E γ 1 h ν hc λ
------ 12 400 eV^ A
λ
λ 12 400 eV^ A
160 keV
= -------------------------------- = 0.0775 A ˙^ ( 0.00775 nm )
λ 2 λ (^1) m^ h e c
--------- 1( – cosθ) hc m (^) e c^2
- = = ------------ 1( – cosθ) =0.0243 1( – cosθ)
λ 2^ hc E 2 γ
-------- 12 400 eV^ A
120 keV
= = -------------------------------- = 0.104 A ˙
( 1 – cosθ) (^ 0.103^ – 0.0775)^ A
cos θ = – 0.049 θ = 92.8°
p γ sin θ = p (^) e sin φ ⇒ p γ c sinθ = p (^) e c sinφ
p (^) e c = E^2 – m 02 c 4 = ( 40 keV +511 keV)^2 – ( 511 keV)^2 = 206 keV
sin θ 120 keV^ ⋅0. 206 keV
φ = 35.6°
- The solar constant (energy/sec/ received from the sun at earth) is . The earth-sun distance is. The radius of the earth is . the radius of Jupiter is and its distance from the sun is . Calculate the average temperature of Jupiter. You may assume that Jupiter is a perfect absorber and that it is in equilibrium with respect to energy received from the sun. What is the most probable wavelength of the EM radiation emitted by Jupiter?
Radiation falls off as
Total energy hitting Jupiter
(b) most probable wavelength
Sun
Earth (^) Jupiter
R J
R E
m^2 W = 1400 W ⁄ m^2 R (^) E =1.5 × 1011 m r (^) E = 2 400, km r (^) J = 71 400, km
R (^) J = 5.2 R (^) E =7.8 × 1011 m
R^2 S (^) jupiter 1400 W ⁄ m^2 *
R E^2
( 5.2 R E )^2
⇒ = ---------------------- =51.8 W ⁄ m^2
P (^) JRec^ 51.8 W m^2
= ------ * π r^2 j
P (^) JRad^ = R 4 π r^2 j^ =σ T^4 4 π r^2 j
51.8 * π r^2 j^ = 4 π r^2 j^ σ T^4
T^4 51.
4 σ
= ---------- T 51.8^ W^ m
( ⁄^2 )
4 * 5.67 * × 10 – 8 W ⁄ m^2 k^4
T = 123 ° K
4.97λ m kT = hc
λ m hc 4.97 kT
----------------- 12 400^ A
, ˙ eV 4.97 * .025 *^123 300
--------- eV
=2.43 10 5 A ˙
= 24.3μ
- In semiconductors, electrons with charge and effective mas can carry current. In
additon, there are excitations, called holes that behave exactly as if they were positively charged electrons with effective mass. In addition semi-conductors have a large dielectric
constant , so the Coulomb force between the electron and the hole is. Electrons and holes can form a “hydrogenic atom” called an exciton. In Germanium, , , and. Calculate the ground state energy and wavelength of the first Balmer line of an exciton in Germanium.
since
m (^) h^ *
κ – ke^2 ⁄ ( κ r^2 )
m (^) e^ *^ = 0.082 m (^) e m (^) h^ *^ = 0.042 m (^) e κ = 16.
μ reduced mass
m (^) h^ * m (^) e^ * m (^) h^ *^ + m (^) e^ *
-------------------- (^ 0.042)^ (^ 0.082)
= = = ------------------------------------- m (^) e = 0.0278 m (^) e
E (^) n^ μ k
(^2) Z (^2) e 4
κ^2 ⋅ 2 n^2 h^2
------------------------- μ m (^) e κ^2
n^2
= = – ----- E (^) B k k κ
E 1 0.
( 16 )^2
= – ----------------× 13.6 ⋅ V =–1.48 × 10 –^3 eV
hc λ B
------ E 3 – E 2 E^1
– -----^5
------ E 1 ⇒ λ B hc 5 36
------ E 1
λ B^1240 ( 5 ⁄ 36 ) ×1.48 × 10 –^3
= ------------------------------------------------- = 6.03 × 106 nm = 6.03 mm
- Helium atoms of mass and kinetic energy are scatteredoff cubic crystalline material. The smallest scattering at which a Bragg peak is observed is at. What is the lattice spacing of the crystal?
Bragg scattering:
M = 3.72Gev/ c^2 K = 1.2 × 10 – 3 eV
Θ = 2 θ ⇒ θ = 70 °
λ hc 2 mc^2 K
[ 2 ×3.72 × 10 9 ×1.2 × 10 –^3 ]
= = -------------------------------------------------------------------------- 1 ⁄ 2 - = 0.415 nm
2 d sin θ = n λ →λ n =
Θ = 2 θ θ d
d λ 2 sinθ
2 sin 70 °
= = -------------------- = 0.22 nm = d
- Calculate the wavelength of light emitted when an electron in a one-dimensional square well of length 3 nm undergoes a transition from level 4 to level 2.
E (^) n n
(^2) h 2
8 m λ^2
------------- n^2 (^ hc )
2
8 mc^2 L^2
------------------- n^2 (^1240 )
2
8 ×0.511 × 106 × 3 2
= = = ---------------------------------------------- = 0.042 eV n n
hc λ
------ = ∆ E = ( 4 2 – 2 2 ) E 1 = 12 ×0.042 eV
λ ∆ E hc
-------^
– 112 ×0.
-------------------------^
