### PhotoElectric Effect 2

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1. The frequency of a photon of light is 2 x 1015Hz. It is shone onto the surface of a metal which has a work function of 3 x 10-19J.

1.1. Calculate the energy of the photon.
E = hf
E = (6,63 x 10-34)(2 x 1015)
E = 1,326 x 10-18J
Audio 1.1

1.2. Calculate the kinetic energy of the ejected electron.
Ek = (1,326 x 10-18) – (3 x 10-19)
Ek = 1,026 x 10-18J
Audio 1.2

1.3. Calculate the speed of a photoelectron.
1.4. The intensity of the light is increased.

1.4.1. What would the effect on the kinetic energy of the ejected electrons?
There would be no effect on the kinetic energy of the photoelectrons.
Audio 1.4.1

1.4.2. What would the effect on the number of ejected electrons?
A greater number of electron would be released since there is more light.
Audio 1.4.2

1.5. How is the speed of the photon affected by a change in its frequency?
No change. Photon speed is unaffected. (photon is the light)
The speed of light is a constant and equal to 3 x 108ms-1 irrespective of its frequency.
Audio 1.5

2. The frequency of a photon of light is 3 x 1015Hz.
It is shone onto nickel metal which has a work function of 8 x 10-19J.

2.1.Calculate the energy of the photon.
E = hf
E = (6,63 x 10-34)(3 x 1015)
E = 1,99 x 10-18J
Audio 2.1

2.2. Calculate the kinetic energy of the ejected electrons.
Ek = photon – WF
Ek = (1,99 x 10-18) – (8 x 10-19J)
Ek = 1,19 x 10-18J
Audio 2.2

3. A metal has a work function of 5 x 10-19J. A photon of light is shone onto the metal and photoelectrons are ejected with a kinetic energy of 2 x 10-19J.

3.1. Calculate the energy of the photon.
E = (5 x 10-19) + (2 x 10-19)
E = 7 x 10-19J
Audio 3.1

3.2. Calculate the frequency of the photon.
E = hf
(7 x 10-19) = (6,63 x 10-34)f
f = 1,06 x 1015Hz
Audi 3.2