How to increase photocurrent (1 Viewer)

[mrpotatoed]

Does increasing the surface area of a solarcell increase photocurrent?

My teacher said increasing the frequency of incident light, and hence the kinetic energy of electrons, does not affect the current. Dot point disagrees. Can anyone shed light on this please.

(unintentional pun)

 

[Fizzy_Cyst]

Does increasing the surface area of a solarcell increase photocurrent?

My teacher said increasing the frequency of incident light, and hence the kinetic energy of electrons, does not affect the current. Dot point disagrees. Can anyone shed light on this please.

(unintentional pun)

Increasing surface area would increase photocurrent if the increased surface area allows extra photons to be absorbed, which we would assume that it does.

Your teacher is correct, the way which we look at it in the HSC, is that frequency has no impact. If you think about it from the view point that if you have 2 light sources, same intensity, but different frequency, they are both causing the same amount of photoelectrons to be ejected from the surface per unit time, hence q/t would be constant, hence current is constant.

 

[mrpotatoed]

For 3.12.5 in phy dot pt, it says:

"Identify two ways the photocurrent from the solar cell could be increased, and justify each way"

The answer of which is:

"Increase the frequency of the incident light. The higher the frequency, the more energy the photon carries (E=hf), so each photon will be released from the surface with greater kinetic energy. More electrons will flow per second, so the current increases"

I guess that kind of makes sense, like if you hypothetically had one electron in a circuit, and it was going at 1m/s in a 5 meter circuit, it would pass a certain point once every 5 seconds, increase the speed (otherwords ke) of the electron to 5m/s, and it passes the point once every second, so the current is 5x greater, right?

But then my teacher and yourself are saying differently

 

[InteGrand]

If you think about it from the view point that if you have 2 light sources, same intensity, but different frequency, they are both causing the same amount of photoelectrons to be ejected from the surface per unit time, hence q/t would be constant, hence current is constant.

If we keep intensity constant, but change the frequency, why doesn't the rate of photoelectron ejection change? If intensity (=power/area) is constant, with area constant, it means the power (= energy/time) is constant. Since energy of a photon = hf, shouldn't increasing f mean the energy per photon increases? Then in order to have Intensity constant, we'd require a lower rate of arrival of photons (since each photon has more energy). This would lead to a lower rate of photoelectron ejection, wouldn't it?

(I'm not saying anything about what the HSC requires us to say; that's a different matter entirely.)

 

[mrpotatoed]

If we keep intensity constant, but change the frequency, why doesn't the rate of photoelectron ejection change? If intensity (=power/area) is constant, with area constant, it means the power (= energy/time) is constant. Since energy of a photon = hf, shouldn't increasing f mean the energy per photon increases? Then in order to have Intensity constant, we'd require a lower rate of arrival of photons (since each photon has more energy). This would lead to a lower rate of photoelectron ejection, wouldn't it?

(I'm not saying anything about what the HSC requires us to say; that's a different matter entirely.)

Oops yeah sorry, I got mixed up. My teacher said photocurrent doesn't increase only when intensity is kept constant since its energy per area per second... so then dot point is correct also. cheers guys

 

[Fizzy_Cyst]

If we keep intensity constant, but change the frequency, why doesn't the rate of photoelectron ejection change? If intensity (=power/area) is constant, with area constant, it means the power (= energy/time) is constant. Since energy of a photon = hf, shouldn't increasing f mean the energy per photon increases? Then in order to have Intensity constant, we'd require a lower rate of arrival of photons (since each photon has more energy). This would lead to a lower rate of photoelectron ejection, wouldn't it?

(I'm not saying anything about what the HSC requires us to say; that's a different matter entirely.)

Please see this part of my post

Your teacher is correct, the way which we look at it in the HSC, is that frequency has no impact.

This is, after all, a HSC site, correct?

#hscsez

 

[sy37]

1. Increasing the surface area for photo reception (more photons can strike)

2. Increasing the intensity of light source (more photons per area)

 

[iforgotmyname]

Well current is defined as the number of electron flowing in circuit and intensity is defined as number of photons.
If frequency is larger than threshold frequency and intensity increases, more electron will be liberated from the lattice. If you increase the frequency you only increase the ke of electrons therefore increasing the voltage of photocurrent.

Increasing area usually increases the photocurrent but that's if all of the extended area of the metal is exposed to the light