Hi, I was wondering if someone could please help me out with the following question:

The answer is C.

As the power is doubled, I thought the energy would double and as E=hf, consequently the KE would double whilst the number of photoelectrons emitted per second stays the same.

However, after rereading the question, it states 'power of the light incident', so does the question imply that the frequency of the light remains the same as the light source is the same, thus the energy of one photon is the same but if power is doubled, then number of photoelectrons per second is doubled?

Thank you very much in advance!

The answer is C.

As the power is doubled, I thought the energy would double and as E=hf, consequently the KE would double whilst the number of photoelectrons emitted per second stays the same.

However, after rereading the question, it states 'power of the light incident', so does the question imply that the frequency of the light remains the same as the light source is the same, thus the energy of one photon is the same but if power is doubled, then number of photoelectrons per second is doubled?

Thank you very much in advance!

The power doubling means the intensity of light doubles (I=P/A). But this will not change the KE because they are independent of each other and increasing the intensity only increases the number of electrons released from the metal's surface.

Hope that helps

Hope that helps

Original post by pinkrocket

The power doubling means the intensity of light doubles (I=P/A). But this will not change the KE because they are independent of each other and increasing the intensity only increases the number of electrons released from the metal's surface.

Hope that helps

Hope that helps

Hi, thank you very much for the response! I wasn't aware of the equation that you mentioned and thus the relationship between power and intensity. May I ask if 'A' in the equation refers to cross sectional area then?

Original post by Felix'sfreckles

Hi, thank you very much for the response! I wasn't aware of the equation that you mentioned and thus the relationship between power and intensity. May I ask if 'A' in the equation refers to cross sectional area then?

Yes, that's right, in this example the area would be the area the light is shining on the metal.

You can get the intensity equation from the definition 'intensity is the power transferred per unit area', therefore intensity = power/area.

Original post by pinkrocket

Yes, that's right, in this example the area would be the area the light is shining on the metal.

You can get the intensity equation from the definition 'intensity is the power transferred per unit area', therefore intensity = power/area.

You can get the intensity equation from the definition 'intensity is the power transferred per unit area', therefore intensity = power/area.

Alright, thank you very much!

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