Question #58653, Physics / Quantum Mechanics | completed
A particular x-ray photon has wavelength of 41.5pm. Calculate photon (a) energy, (b) frequency, (c) momentum.
Solution:
λ=41.5pm=41.5⋅10−12m
a) Relationship between wavelength of photons and its energy: E=hν=h⋅λ′c
where c=3⋅108sm — is the speed of light and
h=6.626070040(81)×10−34J⋅s=4.135667662(25)×10−15eV⋅s−Planck’s constantE=h⋅λc=4.136⋅10−15[eV⋅s]⋅41.5⋅10−12[m]3⋅108[m/s]=29899[eV]
or in SI units:
E=h⋅λc=6.626⋅10−34[J⋅s]⋅41.5⋅10−12[m]3⋅108[m/s]=4.79⋅10−15[J]
b) It is known the frequency of photon f can find from
f=λc=41.5⋅1012[m]3⋅108[m/s]=7.23⋅1018[Hz]
or can find from a):
E=h⋅λc=h⋅f⟹f=hE=6.626⋅10−34[J⋅s]4.79⋅10−15[J]=4.136⋅10−15[eV⋅s]29899[eV]=7.23⋅[Hz]
c) We can calculate the momentum of the photon using the famous De Broglie wavelength formula:
λ=ph⟹p=λh=41.5⋅10−12[m]6.626⋅10−34[J⋅s]=1.597⋅10−23[kg⋅m/s]Answer:
a) E=hλc=29899eV=4.79⋅10−15J
b) f=λc=7.23⋅1018Hz
c) p=λh=1.597⋅10−23kg⋅m/s
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