Physik IV - L osungen - Serie 2 · Physik IV - L osungen - Serie 2 11. M arz 2011 1. 2. (a) Inverse...
Transcript of Physik IV - L osungen - Serie 2 · Physik IV - L osungen - Serie 2 11. M arz 2011 1. 2. (a) Inverse...
Physik IV - Losungen - Serie 2
11. Marz 2011
1
2. (a) Inverse photoelectric effect The total of the kinetic energy ofan incident electron can be converted to radiation, i. e.
hν =hc
λmin
= Ekin → λmin =hc
Ekin
In the non-relativistic case Ekin = mv2/2 and therefore
λmin =2hc
mv2=
8h
mc=
2× 6.6× 10−34
9.11× 10−31 × 2.99× 108= 0.194 A.
In the relativistic case the kinetic energy is given by
Erelkin =
mc2√1− v2/c2
−mc2
and the relativistic cut-off wavelength of the Bremsstrahlung is0.157 A.
(b) A current of 1 mA corresponds to n electrons per seconds each withan energy Erel
kin. The number of electrons is calculated from
I =dQ
dt=Q
t=ne
t
with the electron charge e. Therefore n = It/e = 6.2 · 1015 s−1 andthe power of the electrons hitting the anode is P = nErel
kin = 79 Wwhere we have calculated the kinetic energy Erel
kin = 1.2 · 10−14 Jusing the formula in (a). Take 99% therefrom to get the coolingpower needed to cool the anode during operation.
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