Photoelectric effect question, help , thanks

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SUMMARY

The discussion focuses on calculating the value of h/e using experimental data from a photoelectric effect experiment involving metallic potassium as the photocathode. The stopping voltages measured for light wavelengths of 569 nm and 405 nm are 0.10 volts and 0.99 volts, respectively. The experimental value of h/e calculated is 4.16862E-15 V*s. The accepted value of h/e is 4.14x10-15 V*s, leading to a percentage difference that can be calculated using the formula: 100·(experimental - accepted)/accepted.

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Photoelectric effect question, help please, thanks!

In the standard experiment the stopping voltage is measured for several wavelengths of incident light. In this apparatus the photocathode is metallic potassium. Suppose that the apparatus has not been used for some time and that its calibration is uncertain. When monochromatic light with a wavelength of 569 nm shines on the potassium surface, the photoelectric current is stopped by a retarding voltage of 0.10 volts. When light with a wavelength of 405 nm is used, the stopping potential is 0.99 volts.
a) Determine from these data the value of h/e.
This i already got: 4.16862E-15 V*s

b) How does the value of h/e determined from these data compare to accepted value? Find the percentage difference between the value value from these data and the accepted value. [ 100·(experiment-accepted)/accepted ]
***I'm stuck here, e.g. what's the accepted value if part a is the experimental value, or is it the other way around, thanks: Can anyone please help me with this part of the problem, thanks yall!
 
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Both h (Planck's constant) and e (elementary charge) are constants in nature, so the "accepted value" (maybe you mean expected value?) is just h/e = 6.63x10-34/1.6x10-19 = 4.14x10-15. Compare that to the experimental value using the method that was already given to you.
 
yep, you do exacly that.
 

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