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Physics 14 Online
OpenStudy (eujc21):

8. Find the momentum of a photon in eV/c and in Kg. m/s if the wavelength is (a) 400nm ; (b) 1 Å = 0.1 nm, (c) 3 cm ; and (d) 2 nm .

OpenStudy (eujc21):

\[p=h/\lambda\] where \[h=?\]

OpenStudy (eujc21):

h = plank's constant*

OpenStudy (eujc21):

(a) I got this \[1.656517*10^{-27}(m*kg)/s\]

OpenStudy (eujc21):

for part two on (a) I used \[p=E/c\] where \[E=hc/\lambda\] and I got this answer \[4.9*10^{-19}(eV/c)\] can I get anyone else to confirm this?

OpenStudy (anonymous):

Look at the section on momentum: http://en.wikipedia.org/wiki/Electron-volt It says 1 Gev/c = 5.344 x 10^-19 kg m/s so 1 eV/c = 5.344 x 10^-28 kg m/s Using your answer for the first part of the question, I get p = 3.1 eV/c

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