Asked by Sasha
the first excited state of the nucleus of uranium-235 is 0.051 MeV above the ground state. what is the wavelength of the photon emitted when the nucleus makes a transition to the ground state? (in meters)
Answers
Answered by
drwls
Convert 0.051 MeV to energy in Joules.
1 eV = 1.6*10^-19 J
1 MeV = 1.6*10^-13 J
Call it E.
Then use
(wavelength) = E/(h c)
where h is Planck's constant and c is the speed of light.
1 eV = 1.6*10^-19 J
1 MeV = 1.6*10^-13 J
Call it E.
Then use
(wavelength) = E/(h c)
where h is Planck's constant and c is the speed of light.
Answered by
Prospecs
drwls made one crucial error.
Wavelength equals (h c) / E, because who in their right mind would think that the wavelength is directly proportional to energy?
Wavelength equals (h c) / E, because who in their right mind would think that the wavelength is directly proportional to energy?
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