Year 12 Physics Assignment 13
Section 3-Topic 3
Photons
Solutions
1.
The energy of the photon is
E = hf = (6.6310–34 J · s)(102.1106 Hz) = 6.7710–26 J =
4.2310–7 eV.
2.
We find the wavelength from
 = c/f = hc/E = (1.24103 eV · nm)/(200103 eV) = 6.2210–3 nm.
Significant diffraction occurs when the opening is on the order of the wavelength. Thus there
would be insignificant diffraction
through the doorway.
3.
At the threshold frequency, the kinetic energy of the photoelectrons is zero, so we have
KE = hf – W = 0;
hfmin = W0 ;
(6.6310–34 J · s)fmin = 4.310–19 J, which gives fmin =
6.51014 Hz.
4.
At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so we have
KE = hf – W0 = 0;
hc/max = W0 , or
max = (1.24103 eV · nm)/(3.10 eV) = 400 nm.
5.
The energy of the photon is
E = hf = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(390 nm) = 3.18 eV.
The maximum kinetic energy of the photoelectrons is
KEmax = hf – W0 = 3.18 eV – 2.48 eV =
0.70 eV.
We find the speed from
KEmax = !mv2;
(0.70 eV)(1.6010–19 J/eV) = !(9.1110–31 kg)v2, which gives v =
5.0105 m/s.
6.
The photon of visible light with the maximum energy has the least wavelength:
hfmax = (1.24103 eV · nm)/min = (1.24103 eV · nm)/(400 nm) = 3.10 eV.
Electrons will not be emitted if this energy is less than the work function.
The metals with work functions greater than 3.10 eV are
copper and iron.
7.
(a) At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so we have
KE = hf – W0 = 0;
W0 = hc/max = (1.24103 eV · nm)/(570 nm) =
2.18 eV.
(b) The stopping voltage is the voltage that gives a potential energy change equal to the maximum
kinetic energy:
KEmax = eV0 = hf – W0 ;
(1 e)V0 = [(1.24103 eV · nm)/(400 nm)] – 2.18 eV = 3.10 eV – 2.18 eV = 0.92 eV,
so the stopping voltage is
0.92 V.