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Questions on the Photoelectric Effect

Q6.

(a) The photoelectric effect is represented by the equation hf = Φ+ Ek. What does Ek represent?

Ek represents the maximum kinetic energy of the emitted photoelectrons.

(2 marks)

(b) A metal plate is illuminated with electromagnetic radiation of wavelength 190 nm. The metal has a work function of 7.9 × 10–19 J.

(i) Calculate the frequency of the incident electromagnetic radiation.

c = fλ

∴ f = c/λ

f = 3.0 x 108/(190 x 10-9) need to know your prefixes!

f = 1.58 x 1015

f = 1.6 x 1015 Hz

(ii) Show that the metal plate will emit photoelectrons when illuminated with radiation of this wavelength.

For photoelectrons to be emitted the energy given by each photon must be equal to or greater than the work function.

E = hf = 6.63 x 10-34 x 1.58 x 1015

E = 1.0 x 10-18 J

This is greater than the work function value so photoelectrons will be emitted.

OR

threshold frequency = Φ/h

= 7.9 × 10–19/6.63 x 10-34

= 1.2 x 1015 Hz

The frequency of the incident light is higher than this so photoelectrons will be emitted.

(iii) The radiation incident on the metal plate remains at a constant wavelength of 190 nm but its intensity is now doubled. State and explain the effect this has on the emitted photoelectrons.

Only one photon interacts with each electron. Doubling the intensity of the radiation means that the number of photons per second faling on the metal plate is doubled. Therefore twice as many photoelectrons will be released in a given time.However the (maximum) (kinetic) energy of the released photoelectrons will be constant.

 

(8 marks)

(Total 10 marks)