Physik IV - L osungen - Serie 2 · Physik IV - L osungen - Serie 2 11. M arz 2011 1. 2. (a) Inverse...

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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.

Physik 4 2011 Page 1