Find the threshold frequency for a calcium

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SUMMARY

The threshold frequency for a calcium surface with a work function of 3.33 eV is calculated using the formula f = W/h. The correct conversion of the work function from electron volts to joules is W = 3.33 eV * 1.6 x 10^-19 J/eV, resulting in W = 5.33 x 10^-19 J. Using Planck's constant h = 6.63 x 10^-34 J·s, the threshold frequency is f = 5.02 x 10^14 Hz. The initial calculation of 8.04 x 10^14 Hz was incorrect due to a miscalculation, but the final confirmation shows the book's answer was erroneous, validating the user's calculation.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with the concepts of work function and threshold frequency
  • Knowledge of Planck's constant (h = 6.63 x 10^-34 J·s)
  • Ability to convert electron volts to joules
NEXT STEPS
  • Review the photoelectric effect and its implications in quantum physics
  • Study the calculation of threshold frequency in different materials
  • Learn about the significance of work function in electron emission
  • Explore advanced topics in quantum mechanics related to photon energy
USEFUL FOR

Students studying quantum physics, educators teaching the photoelectric effect, and anyone interested in the properties of materials in relation to electron emission.

crimsondarkn
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Homework Statement



This is a pretty simple problem... Yet I got the wrong answer.. Hmm...

"Find the threshold frequency for a calcium surface whose work function is 3.33eV"

W=3.33eV = 3.33eV * 1.6x10^-19 J = 5.33x10^19 J
f = ?
h=6.63x10^-34 J*S

Homework Equations



f=W/h

The Attempt at a Solution



f=(3.33eV)(1.6x10^-19 J) / (6.63x10^-34 J*S) = 8.04x10^14 Hz <- My answerThe answer in book is 5.02 x10^14 Hz.

Can anyone tell me what I'm doing wrong?
 
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turns out the book was wrong... my answer was right. yah...
 

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