Photoelectric effect: (2 Viewers)

[umm what]

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[someth1ng]

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1) Why experiments are carried out in a vacuum? (1 mark)
2) Why must the surface of the test metal be carefully prepared? (1 mark)
3) Identify the relo b/w Planks Constant, the work function for a particular metal and the KE of the photoelectrons. (1 mark)
4) What would be some controlled variables in the experiment? (1 mark)
I wrote surface of the metal....any others?

Are you sure this is photoelectric effect? Sounds more like cathode rays for #1

1.
It must be carried out in a vacuum because the presence of gas will cause collisions because the molecules of gas and the electrons will collide causing difficulty of observation of the photoelectric effect.

2.
Impurities in a metal may cause the work function to change at particular parts of the metal such that some photoelectrons may be emitted even it appears that energy of photons is still below the work function.

3.
The work function is the amount of energy required to liberate a single electron by absorbing the energy of one photon per electron.
The energy of a photon is given by E=hf where h is the Planck's constant.
The energy of the photon must be higher than the work function of a particular metal (W) for the photoelectric effect to occur - if energy of light is below the work function, no electrons will be liberated.
.'. for the photoelectric effect to occur: hf > W
Any excess energy of a photon above the work function will be converted into kinetic energy of the photoelectrons.
.'. KE = hf-W

4.
Controlled variables will depend on what you are looking for? You give me no details about your aim.

 

[umm what]

Are you sure this is photoelectric effect? Sounds more like cathode rays for #1

1.
It must be carried out in a vacuum because the presence of gas will cause collisions because the molecules of gas and the electrons will collide causing difficulty of observation of the photoelectric effect.

2.
Impurities in a metal may cause the work function to change at particular parts of the metal such that some photoelectrons may be emitted even it appears that energy of photons is still below the work function.

3.
The work function is the amount of energy required to liberate a single electron by absorbing the energy of one photon per electron.
The energy of a photon is given by E=hf where h is the Planck's constant.
The energy of the photon must be higher than the work function of a particular metal (W) for the photoelectric effect to occur - if energy of light is below the work function, no electrons will be liberated.
.'. for the photoelectric effect to occur: hf > W
Any excess energy of a photon above the work function will be converted into kinetic energy of the photoelectrons.
.'. KE = hf-W

4.
Controlled variables will depend on what you are looking for? You give me no details about your aim.

ok for no. 4, we need to state the controlled variables for the photoelectric effect.

 

[Timske]

For 1 it is the high density of air molecules

 

[Timske]

Controlled variables is the intensity of light and emr

 

[umm what]

Controlled variables is the intensity of light and emr

and independent variable is frequecny and dependent variable is Ek right?

 

[umm what]

One last question, it asks:

Should the the surface of metal be positively or negatively charged to investigate proper results? (1 mark)

 

[someth1ng]

Should the the surface of metal be positively or negatively charged to investigate proper results? (1 mark)

Assuming the aim is to observe the photoelectric effect, the metal should be negatively charged.

 

[umm what]

Assuming the aim is to observe the photoelectric effect, the metal should be negatively charged.

question asks 'why' ?

 

[someth1ng]

question asks 'why' ?

The question you gave doesn't ask for why the plate should be negative.

Anyway, if it is negatively charged, it will more readily give up charge because the negative charge indicates that there are more electrons than normal allowing them to be liberated. Also, the electric field created by the negative charge promotes the photoelectric effect by repelling electrons away from the plate.

 

[umm what]

Am I correct? esp the last ques. 7

Thanks

 

[umm what]

bump!!!!

 

[barbernator]

u just word for word copied the other dudes response, wtf?