Lösung: Q-0016

Photoeffekt

Innerer Photoeffekt:
a) eU = hf = hc/λ
=> f = eU/h = 5,8E14 Hz
=> λ=hc/eU = 518 nm

b) U_rot=hc/λe = 1,97 V
analog:
U_gelb= 2,18 V
U_grün= 2,34 V
U_kin = 2,64 V

Summe aller Spannungen: 45,7V

Äußerer Photoeffekt:
a) W_kin=W_ph-W_A = hc/λ-W_A = 3,46E-19J
v = $\sqrt{\frac{2W_{\mathrm{kin}}}{m_e}}$= 872489 m/s
b) W_ph=W_A
=> λ=$\frac{\mathrm{hc}}{W_A}$= 319nm
c) W_ph=hc/λ= 7,95E-19 J
W_g= Ue =W_kin = W_ph - W_A
=> U = (W_ph - W_A)/e = 1,07 V

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