How Does Reducing Wavelength Affect Photoelectron Kinetic Energy and Current?

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Homework Help Overview

The discussion revolves around the photoelectric effect, specifically examining how reducing the wavelength of incident radiation affects the maximum kinetic energy of photoelectrons and the resulting photoelectric current. The original poster presents a scenario where the wavelength is decreased while maintaining constant intensity, leading to questions about the implications for kinetic energy and current.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between wavelength, photon energy, and photoelectric current, questioning how frequency impacts the number of emitted electrons and current. Some participants clarify the definition of intensity and its relation to the number of photons, while others explore the implications of changing wavelength on current.

Discussion Status

The discussion is active, with participants providing clarifications about the definitions of intensity and its relationship to photon emission. There is an ongoing exploration of how changes in wavelength and frequency affect the photoelectric current, with no explicit consensus reached yet.

Contextual Notes

Participants are navigating the complexities of the photoelectric effect, particularly how intensity is defined in relation to wavelength and frequency. There is a focus on the implications of maintaining constant intensity while changing wavelength, which raises questions about the number of photons and their effect on current.

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Homework Statement


Radiation of wavelength 240nm gives rise to a maximum photoelectric current, I. The intensity of the incident radiation is maintained constant and the wavelength is now reduced.
State and explain the effect of this change on
(i) The maximum kinetic energy of the photoelectrons

(ii) The maximum photoelectric current I.




The Attempt at a Solution



I understand that for (i) the maximum KE will increase as the wavelength is reduced causing the photon energy to be larger and the maximum KE is independent of the intensity. However, I don't get (ii). My initial answer is that the maximum photoelectric current, I remains constant as the intensity if unchanged. However, the answer says that there are fewer photons per unit time, so the maximum current is smaller. Does frequency also affect the number of electrons emitted per unit time (since frequency is the number of cycles per unit time) and hence affect the magnitude of the photoelectric current?

I hope you can clear up my confusion! Thanks in advance! :)
 
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What do you think the intensity of light means? :)

ehild
 
ehild said:
What do you think the intensity of light means? :)

ehild

Hi there!
Intensity just means the number of photons contained in the light beam. So the higher the intensity, the more photons are contained. I know clearly that if frequency is kept constant, an increase in intensity will increase the photoelectric current because current is the rate of flow of charge and an increase in intensity will increase the number of electrons emitted from the surface. However, in this case, the wavelength of the light beam is reduced, which means that the frequency is reduced as well. How does this affect the intensity and how does this then, affect the maximum photoelectric current?

I hope you can shed some light on this. Thanks!
 
jiayingsim123 said:
Hi there!
Intensity just means the number of photons contained in the light beam.

The intensity of light is the energy transferred in unit time through unit area.

ehild
 
You forgot that intensity of light is also associated with the frequency by the following relation:

I=nhf/A (n=no. of photons emitted per unit time, h=Planck's constant, f=frequency of light, A=area perpendicular to light's direction of propagation)

In this question, it is mentioned that the wavelength is reduced without the intensity changing. In order to effect this, you have to decrease the rate of emission of photons, by virtue of the formula written about.

A closer examination would require you to analyse the effects the increased maximum kinetic energy has on the photoelectric current (I=nAve). However, if you do the mathematics, you would get the same conclusion.

I hope this helps.
 

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