Photoelectric Materials Homework: Calculating Voltage for Extinguishing Current

[jeffy]

Homework Statement

A multi-akali alloy is used as the photocathode material in a photoemissive electron tube. The work function is 1.6eV a blue light of wavelength 420nm is incident to the photocathode. What should be the voltage and polarity to extinguish the current

Homework Equations

kintetic energy=hf-hf0

The Attempt at a Solution

To my understanding a negative voltage with the same magnitude as the work function is needed to extinguish the current completely, and therefore the voltage should be -1.6eV, however,the solution says that the voltage needed to extinguish current is -1.36eV which is the kinetic energy

thanks for reading

 

[KataKoniK]

and responding to this forum post. I can confirm that your understanding is correct. A negative voltage with the same magnitude as the work function is needed to extinguish the current completely. The solution provided may be referring to the kinetic energy of the electrons, which is equal to the difference between the energy of the incident photon (hf) and the work function (hf0). In this case, the voltage needed to extinguish the current would be equal to the kinetic energy, which is -1.36eV. This may have been a typo or a misunderstanding on the part of the solution provider. It is always important to double check equations and solutions to ensure accuracy in scientific calculations.