Describe the de Broglie hypothesis.

An electron is accelerated from rest through a potential difference of 5.0 kV.

(i) Calculate the momentum of the electron after acceleration.
(ii) Calculate the wavelength of the electron.
(iii) Determine the energy of a photon that has the same wavelength as the electron in (b)(ii).

The momentum of the electron is known precisely. Deduce that all the information on its position is lost.

With reference to Schrödinger’s model, state the meaning of the amplitude of the wavefunction for the electron.

all particles have an associated wavelength/behave like waves;
with \(\lambda  = \frac{h}{p}\) and symbols defined/described using terms;

(i) \(p = \left( {\sqrt {2mE}  = \sqrt {2meV}  = } \right)\sqrt {2 \times 9.11 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 5.0 \times {{10}^3}} \);
\( = 3.8 \times {10^{ - 23}}\left( {{\rm{Ns}}} \right)\);

or

\(v = \left( {\sqrt {\frac{{2eV}}{m}}  = } \right)\sqrt {\frac{{2 \times 1.6 \times {{10}^{ - 19}} \times 5.0 \times {{10}^3}}}{{9.11 \times {{10}^{ - 31}}}}} \);

\(p = (mv = )3.8 \times {10^{ - 23}}(Ns)\);

(ii) \(\lambda  = \left( {\frac{h}{p} = } \right)\frac{{6.63 \times {{10}^{ - 34}}}}{{3.8 \times {{10}^{ - 23}}}}\);
=1.7×10^-11m;
This is a question testing units for this option. Do not award second marking point for an incorrect or missing unit.

(iii) \(E = \left( {hf = \frac{{hc}}{\lambda } = } \right)\frac{{6.63 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}{{1.7 \times {{10}^{ - 11}}}}\);
E=1.2×10^-14(J);

or

\(E = (cp = )3.0 \times {10^8} \times 3.8 \times {10^{ - 23}}\);

\(E = 1.2 \times {10^{ - 14}}(J)\);
Allow ECF from (b)(ii).

reference to the Heisenberg uncertainty principle / \(\Delta x\Delta p \ge \frac{h}{{4\pi }}\);
Δp = 0 implies Δx is large /Δx=∞;

the (square of the) amplitude gives the probability of finding the electron at a given point in space;