How Does Changing the Work Function Affect Photoelectron Emission?

AI Thread Summary
The discussion revolves around calculating various parameters related to photoelectron emission from a metal surface with different work functions. For a work function of 3 eV and a maximum speed of ejected electrons at 0.6 x 10^6 m/s, the wavelength of incident photons was incorrectly calculated, suggesting a frequency that is unrealistic for photoelectric effects. The stopping potential was determined to be approximately 1.02 V. When the work function is increased to 5.5 eV, the maximum speed of ejected electrons will decrease, reflecting the energy required to overcome the higher work function. The impact of increasing the intensity of incident light on photo current was also questioned, indicating a need for clarity on the relationship between light intensity and photoelectron emission.
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Modern Physics, homework, please help:(

A metal surface with a work function of (W1=3eV) is illuminated with a monochromatic ultraviolet light of frequency v*1. The maximum speed of the ejected photo electrons was found to be v1=0.6x10^6m/s.

a) compute the wavelength λ of the incident photons.
b) what is the stopping potential?
c) now the metal is in the photo tube is changed so the work function is W2=5.5eV. Everything else remains the same. Compute the maxiumum speed of the ejected photo electrons and discuss.
d) How would the magnitude of the photo current be changed if the intensity of the incident light were increased? Increases, decreases or does not change? ( answer these question for parts (a) and (c)).
 
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You need to post your attempt. What would your relevant equations be?
 


rock.freak667 said:
You need to post your attempt. What would your relevant equations be?

sorry, i didnt know this rule.. actually i knew formulas but i don't know where i should use them. Also this question is from my exam. I couldn't do it there:s

a ) A metal surface with work function φ = 3.0 eV
maximum speed ejected from photo electrons is v_1 = 0.6 *10^6 m/s

K.E _max = h f - φ
1/2 mv^2 = hf - φ
1/2 * 9.1 *10^-31 kg ( 0.6 *10^6 m/s)^2 = 6.625*10^-34 J.s * f - 3.0 * 1.6 *10^-19 J /eV
f = 9.71*10^-5 Hz
wavelength λ = c
= 3 *10^8 m/s / 9.71*10^-5 Hz
= 3.08*10^12 m

b ) stopping potential is K.E = q ΔV
1/2 * 9.1 *10^-31 kg ( 0.6 *10^6 m/s)^2 = 1.6 *10^-19 C * ΔV
ΔV = 1.02V

c) same formula with a
d) i don't know:s

am I right?
 
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You might want to revisit your frequency calculation. The method and numbers used look okay, but your result seems to be out of whack; your value implies a period for the wave of nearly three hours!

For photoelectric emission on metals you'd expect the light to be in the hundreds of terahertz range.
 
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