Light is incident on a metal surface A. A potential difference is applied between A and an electrode B. Photoelectrons arrive at B and the resulting current is measured by a sensitive ammeter. (Note: the complete electrical circuit is not shown.)

(i) The frequency of the light is reduced until the current measured by the ammeter falls to zero. Explain how Einstein’s photoelectric theory accounts for this observation.

(ii) A different metal surface is used so that a current is again measured. Outline the effect on the photoelectric current when the intensity of the light is doubled and the frequency remains constant.

A photon of energy 6.6×10^–19J is incident upon a clean sodium surface. The work function of sodium is 3.7×10^–19J. The photon causes an electron to be emitted from the surface with the maximum possible kinetic energy. The position of this electron is measured with an uncertainty of 5.0×10^–9m.

Calculate the

(i) momentum of the electron.

(ii) uncertainty in the momentum of the electron.

(i) light consists of photons/quanta/packets of energy;
(each) photon has energy E=hf / photon energy depends on frequency;
a single photon interacts with a single electron giving up all its energy;
a certain amount of energy is required to eject an electron from the metal;
if photon energy is less than this energy/work function/frequency below threshold, no electrons are emitted;

(ii) increasing the intensity increases the photoelectric current;
photocurrent will change as a different metal has a different work function/threshold frequency;

(i) \({E_K} = \left[ {6.6 - 3.7} \right] \times {10^{ - 19}} = 2.9 \times {10^{ - 19}}\left( {\rm{J}} \right)\);
\({E_K} = \frac{{p2}}{{2m}} \Rightarrow p = \sqrt {2m{E_K}}  = \sqrt {2 \times 9.1\left( 1 \right) \times {{10}^{ - 31}} \times 2.9 \times {{10}^{ - 19}}} \);
p=7.2×10^-25(kg ms^-1); (allow answers in the range of 7.2 to 7.3×10^-25)
Award [3] for a bald correct answer.

(ii) \(\Delta p = \frac{{\frac{h}{{4\pi }}}}{{\Delta x}} = \frac{{\frac{{6.6\left( 3 \right) \times {{10}^{ - 34}}}}{{4\pi }}}}{{5.0 \times {{10}^{ - 9}}}}\);
=1.1×10^-26(kg ms^-1);
Award [2] for a bald correct answer.