This question is about the photoelectric effect and the de Broglie hypothesis.
When photons are incident on a lithium surface photoelectrons are emitted. The work function φ of lithium is 2.9 eV.
a.Define work function.[1]
▶️Answer/Explanation
Markscheme
a.
minimum energy/work required to remove an electron (from the surface of the substance);
\({f_{\min }} = \frac{{2.9 \times 1.6 \times {{10}^{ – 19}}}}{{6.63 \times {{10}^{ – 34}}}} = \left( {7.0 \times {{10}^{14}}} \right)\left( {{\text{Hz}}} \right)\);
\({\lambda _{\max }} = \left( {\frac{{3.00 \times {{10}^8}}}{{7.0 \times {{10}^{14}}}}} \right) = 4.3 \times {10^{ – 7}}\left( {\text{m}} \right)\);
\(p = \left( {\frac{h}{{{\lambda _{\max }}}} = } \right)\frac{{6.63 \times {{10}^{ – 34}}}}{{4.3 \times {{10}^{ – 7}}}}\);
=1.5×10–27 (kg ms–1);
Allow ECF from (b).
or
\(p = \left( {\frac{\phi }{c} = } \right)\frac{{2.9 \times 1.6 \times {{10}^{ – 19}}}}{{3.00 \times {{10}^8}}}\)
=1.5×10–27 (kg ms–1);
Allow ECF from (b).
This question is about the photoelectric effect.
In a photoelectric experiment, light of wavelength 450 nm is incident on a sodium surface. The work function for sodium is 2.4 eV.
a.(i) Calculate, in eV, the maximum kinetic energy of the emitted electrons.
(ii) The number of electrons leaving the sodium surface per second is 2 \( \times \) 1015. Calculate the current leaving the sodium surface.[5]
▶️Answer/Explanation
Markscheme
a.
(i) photon energy \( = \left( {\frac{{hc}}{\lambda } = \frac{{6.63 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}}{{450 \times {{10}^{ – 9}}}} = } \right)4.4 \times {10^{ – 19}}\left( {\rm{J}} \right)\);
=2.76(eV) ;
2.76 – 2.4 = 0.36 (eV) ;
Award [3] for a bald correct answer.
Award [1 max] if the energy of the photon is not converted from Joules to eV (giving EK =-2.4 eV ).
(ii) 2×1015×1.6×10–19 ;
3×10–4 (A);
Award [2] for a bald correct answer.
light consists of photons;
frequency of photons increases so energy of photons increases;
same intensity of radiation means fewer photons;
fewer photons means fewer (photo)electrons;
(the emitted number of electrons falls)
Examiners report
a.
In part (a)(i) many correct answers were seen, however there were quite a few mistakes in converting the photon energy to eV. Part (a)(ii) was easy, but some candidates thought that the KE of the electrons needed to be used.
Candidates found (b) difficult as the answer is slightly counter-intuitive. Very few candidates seemed to know that equal intensity of light means equal total photon energy per second. Many assumed it meant equal number of photons per second. The answers were mostly disorganised and did not reflect a logical scientific argument that would be expected at this level.