Wavelenght of the incident light

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Homework Help Overview

The discussion revolves around calculating the wavelength of incident light based on the work function of a mercury plate and the observed stopping voltage. The subject area is primarily focused on the photoelectric effect and related quantum mechanics principles.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the photoelectric equation but questions their calculation steps and the values used for constants. Some participants question the value of Planck's constant used in the calculations, suggesting a potential error in the original poster's approach.

Discussion Status

Participants are actively engaging in clarifying the calculations and constants involved. There is a focus on verifying the correctness of the values used and the steps taken, with some participants providing alternative calculations that lead to different results. No consensus has been reached regarding the correct approach or final answer.

Contextual Notes

There is a mention of following an example from class, indicating that the original poster may be relying on instructional material. The discussion also highlights potential confusion regarding the constants used in the calculations.

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Homework Statement


An incident light shines on a mercury metal plate with a work function of -4.5 eV. The stopping voltage is observed to be -1.65 volts. What is the wavelength of the incident light?


Homework Equations


wavelength=hc/E-0(it's a Greek sign n i don't have it in my computer)


The Attempt at a Solution


(4.14 X 10^-15 eVs) (3 X 10^8 m/s) / 1.65 + 4.5 = 1.242 X 10^-6 / 6.15

1.242(10^2)/6.15 = 20.1954

but the right answer is 201.7561 nm


so can anyone please tell me what did I do wrong?
 
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\frac{hc}{\lambda}=\phi +eV_s

Did you use the correct value for h in eV?
 
i used h=4.14 X 10^ -15 am i supposed to use the other one? h=6.626 X 10^-34 ?
 
How'd you go from 1.242 X 10^-6 / 6.15 to 1.242(10^2)/6.15 = 20.1954?

Because if you work out 1.242 X 10^-6 / 6.15 you'd get your answer.
 
I was following one of the examples that my teacher did in class.

well if i do 1.242 X 10^-6 / 6.15...it equals 2.01951 but the right answer is 201.7561 nm
 
1.242 X 10^-6 / 6.15 =201nm
 

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