(a)
When monochromatic light is incident on a particular metal plate, electrons are emitted. The intensity
of the light is then increased.
Explain
(i)
why the maximum kinetic energy of the emitted electrons does not change,
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(ii)
why the number of electrons emitted per second increases.
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(3)
(b)
A potassium metal plate is illuminated with incident light of wavelength 5.10 × 10–7 m. The work
function of potassium is 3.5 8 × 10–19 J.
(i)
Show that the frequency of the incident light is approximately 6 × 1014 Hz.
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(ii)
Calculate the energy, in J, of an incident photon.
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(iii)
Calculate the maximum kinetic energy, in J, of an emitted electron.
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(iv)
The table gives the work function of four metals.
metal
work function / J × 10–19
caesium
3.04
silver
7.57
sodium
3.94
tungsten
7.33
Which of these metals would emit electrons when illuminated with light of wavelength 5.10 ×
10–7 m?
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(7)
(Total 10 marks)
(a)
(b)
(i)
the energy of a photon does not depend on the intensity (1)
so electron gains no extra energy
[or the energy is dependent on the wavelength/frequency] (1)
(ii)
the intensity of the light determines the number of photons
per second (1)
one photon interacts with one electron
[or hence more interactions with electrons] (1)
(i)
3
8
(use of c = f gives) f = 3.00  10 (1)
5.10  10 7
= 5.88  1014 (Hz) (1)
(ii)
(use of E = hf gives) E = 6.63  10–34  5.88  1014 (1)
(allow C.E. for value of f from (i))
= 3.9(0)  10–19 (J) (1)
(iii)
(use of hf =   Ek gives) 3.9  10–19 = 3.58  10–19 + Ek (1)
(allow C.E. for value of E from (ii))
Ek = 3.2  10–20 (J) (1)
(iv)
caesium (1)
(allow C.E. for value of E from (ii))
7
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