Photoelectric Effect difficulties

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SUMMARY

The discussion centers on calculating the maximum kinetic energy (KEmax) of emitted electrons when light strikes a metal surface at double its threshold frequency of 5.6 × 1014 Hz. The relevant equations include E=hv for photon energy and W=hFo for work function. The solution involves determining the energy of the incoming photons and subtracting the work function to find KEmax. The confusion arises from understanding how energy is distributed between freeing the electron and its kinetic energy.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with the equations E=hv and W=hFo
  • Knowledge of threshold frequency and its significance
  • Basic algebra for manipulating equations
NEXT STEPS
  • Calculate the maximum kinetic energy using KEmax = E - W
  • Explore the concept of threshold frequency in different metals
  • Investigate the implications of photon energy on electron emission
  • Review AQA Physics Exam questions related to the photoelectric effect
USEFUL FOR

Students studying physics, particularly those preparing for AQA exams, and educators looking to clarify concepts related to the photoelectric effect and energy calculations.

ysidfa
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Homework Statement


"The threshold frequency of a particular metal surface is 5.6 × 1014 Hz. Calculate the maximum kinetic energy of emitted electrons if the frequency of the light striking the metal surface is double the threshold frequency."
Taken from AQA Physics Exam 2009

Homework Equations


E=hv, W=hFo


The Attempt at a Solution


I tried working out double the threshold freq. and then going into finding the KEmax from there. Got Really confuzed... :(
 
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ysidfa said:

Homework Statement


"The threshold frequency of a particular metal surface is 5.6 × 1014 Hz. Calculate the maximum kinetic energy of emitted electrons if the frequency of the light striking the metal surface is double the threshold frequency."
Taken from AQA Physics Exam 2009

Homework Equations


E=hv, W=hFo

The Attempt at a Solution


I tried working out double the threshold freq. and then going into finding the KEmax from there. Got Really confuzed... :(
Assume the photon with double the threshold frequency gives all its energy to the electron. How much is used in freeing the electron from the surface? What happens to the rest of the energy?

AM
 

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