Q#4 (Past Exam Paper – November 2017 Paper 42 Q10)
(a) A metal surface is illuminated with light of a single wavelength λ.
On Fig. 10.1, sketch the variation with λ of the maximum kinetic energy EMAX of the electrons emitted from the surface.
On your graph mark, with the symbol λ0, the threshold wavelength.
Fig. 10.1
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(b) A neutron is moving in a straight line with momentum p.
The de Broglie wavelength associated with this neutron is λ.
On Fig. 10.2, sketch the variation with momentum p of the de Broglie wavelength λ.
Fig. 10.2
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Solution:
(a)
λ0 marked and graph line passing through EMAX = 0 at λ = λ0
graph line with λ always < λ0
negative gradient with correct concave curvature
{Energy of photon: E = hf = hc / λ
Energy from photon = Work function + KE
Energy is required for the electrons to break free from the metal. The remaining energy goes as kinetic energy.
The minimum energy required to liberate an electron is called the work function which occurs at a threshold frequency f0 and corresponding threshold wavelength λ0 (this is the max wavelength for emission. Photons of greater wavelength will not cause emission).
If the energy of the photon is equal to the work function, no energy goes as KE.
→ So, when λ = λ0, EMAX = 0.
The threshold wavelength λ0 is the maximum wavelength for photoemission. Photons of greater wavelength will not cause emission).
→ So, the line is always in the region < λ0.
Energy of photon: E = hf = hc / λ
KE or EMAX = Energy of photon – Work function
OR EMAX = hc / λ (– Work function)
The EMAX is inversely proportional to λ. The graph is that of an inverse relationship, having a negative gradient.}
(b)
curve with negative gradient and correct concave curvature
not touching either axis
The de Broglie wavelength and the momentum are related by the equation:
λ = h / p
This is an inverse relationship graph. The graph does not touch either axis.
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