Work function of silver given wavelength

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The discussion revolves around calculating the work function of silver using the wavelength of light that can eject electrons from its surface. The wavelength provided is 262 nm, leading to a calculated frequency of 1.15 x 10^15 Hz. The initial work function calculation yields 7.60 x 10^-19 J, which is incorrect as the expected value is around 4.74 eV. The error is resolved by converting the energy from joules to electron volts, resulting in the correct work function. This highlights the importance of unit conversion in physics problems.
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Homework Statement



If electrons can be ejected from a silver surface using light with wavelengths as large as 262 nm. What is the work function for silver?

Homework Equations



work function=hf
c=f\lambda

The Attempt at a Solution



Given the wavelength, I solved for f = (3x108)/(262x10-9) = 1.15x1015

Then times by Planck's constant to get work function=(6.63x10-34)x(1.15x1015) gives me 7.60x10-19

...which is completely silly because the answer is supposed to be near 4.74
 
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This answer is 7.60x10-19 J, usually work function is in eV.

Divide 7.60x10-19 by 1.6(10-19) to convert it to eV.
 
OH! Haha, thank you!
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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