Calculate the wavelength of the light.

Electrons emitted from the surface of the photocell have almost no kinetic energy. Explain why this does not contradict the law of conservation of energy.

Radiation of photon energy 5.2 x 10^–19 J is now incident on the photocell. Calculate the maximum velocity of the emitted electrons.

Describe the change in the number of photons per second incident on the surface of the photocell.

State and explain the effect on the maximum photoelectric current as a result of increasing the photon energy in this way.

wavelength = «\(\frac{{hc}}{E} = \frac{{1.99 \times {{10}^{ - 25}}}}{{{\text{3}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 19}}}} = \)» 5.7 x 10^–7 «m»

If no unit assume m.

«potential» energy is required to leave surface

Do not allow reference to “binding energy”.
Ignore statements of conservation of energy.

all/most energy given to potential «so none left for kinetic energy»

energy surplus = 1.7 x 10^–19 J

v_max = \(\sqrt {\frac{{2 \times 1.7 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}}  = 6.1 \times {10^5}\) «m s^–1»

Award [1 max] if surplus of 5.2 x 10^–19J used (answer: 1.1 x 10⁶ m s^–1)

«same intensity of radiation so same total energy delivered per square metre per second» 

light has higher photon energy so fewer photons incident per second

Reason is required

1:1 correspondence between photon and electron

so fewer electrons per second

current smaller

Allow ECF from (c)(i)
Allow ECF from MP2 to MP3.