What is the work function for silver?

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Homework Help Overview

The discussion revolves around the work function of silver in the context of the photoelectric effect, specifically focusing on calculations involving the cut-off frequency and the maximum kinetic energy of ejected electrons when exposed to ultraviolet light.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to calculate the work function using the relationship between frequency and energy, while questioning the appropriate formulas to apply. Some are exploring the connection between the cut-off frequency and maximum kinetic energy, raising concerns about the variables involved in the equations.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and expressing confusion about the equations and variables needed to find the maximum kinetic energy. Some guidance has been offered regarding the use of known constants and the photoelectric equation, but no consensus has been reached on the approach to take.

Contextual Notes

Participants are working under the constraints of needing to derive values from given wavelengths and frequencies, while navigating the complexities of the photoelectric effect equations. There is a noted lack of clarity regarding the relationship between cut-off potential and kinetic energy.

Jas014
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The wavelength associated with the cut-off frequency, for silver is 325 nm.

a) What is the work function for silver?
b) Find the maximum kinetic energy and speed of the electrons ejected from a silver surface by ultraviolet light of wavelength 254 nm.

So far for part a) i have:

f = c / 325 = 9.25 * 10 ^ 14

After that I am not to sure if I am suppose to use this formula

9.25 * 10^14 = hc/325e - work function
 
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The equation describing the photoelectric effect is:

hf=E_{max}+\phi where \phi is the work function.

Now, what is E_max for the case of the cutoff frequency?
 
im having the same problem where I am supposed to find its max. kinetic energy of the photoelectron in electron volts(eV) and joules(J), but I am completely stuck.

if i use hf = Emax + work, i have planks constant (h) and frequency, but there are two variables left to find...
 
sodr2 said:
im having the same problem where I am supposed to find its max. kinetic energy of the photoelectron in electron volts(eV) and joules(J), but I am completely stuck.

if i use hf = Emax + work, i have planks constant (h) and frequency, but there are two variables left to find...
Go through the link that G01 has given. Jot down the data given to you and make the appropriate substitution in the photoelectric equation and obtain the KE. h, m_e are known constants.
 
i don't see anything in that link... for my problem i have the wavelength, cutoff potential and the frequency, I am only confused on the equations to reach to the maximun Ek of the ejected electrons in electron volts(eV) and joules(J).
 
Well then how is the cut-off potential related to the kinetic energy? can you determine it from the given quantities?
 

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