Problems:
- SM&M 2-9.
Photons/sec = Energy/sec / Energy/photon
= Power / hf
= 1.e5 watts / (6.6e-34 joules sec 9.4e7 /sec)
= 1.6e30 photons/sec
How far away from the antenna would you have
to go to have a photon intensity of 3x10^16 photons/cm^2/sec (the equivalent
of a photon density of 10^6 /cm^3).
Solid angle = Area/R^2 = (.01 m)^2/R^2 = .001/R^2
% of photons in Solid Angle = .001/ Pi R^2
= 3e16/1.6e30 = 1.86e-14
----> R = Sqrt[.001 / (Pi 1.86e-14)] = 1.31e5 m = 131 km
You can typically detect an FM radio station
50 miles away.
Given that, what is the intensity of photons/cm^2 necessary
for a radio receiver to detect a signal?
photons/sec on 1cm^2 = .001 m^2/Pi/(8e4 m)^2 *1.6e30 photons/sec
= 8e16 photons/sec on 1 cm^2
- SM&M 2-16
E = h f = h c / lambda = 1241 eV nm/ 300 nm
= 4.14 eV
---> Lithium and Beryllium would emit electrons.
KE = E_gamma - Work function
= 4.14 eV - 2.3 eV = 1.84 eV - Lithium
= 4.14 eV - 3.9 eV = .24 eV - Beryllium
- SM&M 3-28
Z_Chromium = 24
E_n = -13.6 Z^2/n^2 = -7.83 keV/n^2
E_1i = initial energy of electron 1 = -7.83 keV/2^2 = -1.96 eV
E_1f = final energy of electron 1 = -7.83 keV/1^2 = -7.83 keV
E_2i = initial energy of electron 2 = -7.83 eV/4^2 = -.49 keV
E_2f = final energy of electron 2 = KE
Energy Conservation:
E_1i + E_2i = E_1f + E_2f
-1.96 -.49 = -7.83 +KE
---> KE = 5.41 keV
- SM&M 4-12
You can do this problem either relativistically or non-relativistically:
pc = h c /lambda = 1241 eV nm / .1 nm = 12410 eV
Relativistically:
E = Sqrt[(pc)^2 + (mc^2)^2] = Sqrt[(.511e6)^2 + (1.241e4)^2]
= 511151 eV total energy ----> 151 eV kinetic energy.
Non-relativistically:
KE = p^2/2 m = (pc)^2 / 2 mc^2 = (1.241e4)^2/2/.511e6
= 150 eV
- SM&M 4-28
dr = r
dp dr = hbar/2 --> dp = hbar/2r
KE = dp^2/2m = hbar^2/(8 m r^2)
U = -k e^2/r
E = KE + U = hbar^2/(8 m r^2) - k e^2/r
dE hbar^2 k e^2
0 = ---- = - ------- + -----
dr 4 m r^3 r^2
---> r = hbar^2/(4 m k e^2) = a_0/4 = .0529 nm/4 = .0132 nm
E = 4 E_0 = 4 * 13.6 eV
Not too bad for a 1-dimensional model!