a) The frequency of the photon is [tex]7.16\cdot 10^{20}Hz[/tex]
b) The wavelength of the photon is [tex]4.19\cdot 10^{-13} m[/tex]
c) The wavelength of the photon is about 100 times larger than the nuclear radius
Explanation:
a)
The energy of a photon is given by
[tex]E=hf[/tex] (1)
where:
[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant
f is the frequency of the photon
The photon in this problem has an energy of
[tex]E=2.5 MeV = 2.5\cdot 10^6 eV[/tex]
And keeping in mind that
[tex]1eV = 1.6\cdot 10^{-19} J[/tex]
we can convert to Joules:
[tex]E=(2.5\cdot 10^6)(1.9\cdot 10^{-19})=4.75\cdot 10^{-13} J[/tex]
And now we can use eq.(1) to find the frequency of the photon:
[tex]f=\frac{E}{h}=\frac{4.75\cdot 10^{-13}}{6.63\cdot 10^{-34}}=7.16\cdot 10^{20}Hz[/tex]
b)
The wavelength of a photon is related to its frequency by the equation
[tex]c=f\lambda[/tex]
where
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the photon in this problem,
[tex]f=7.16\cdot 10^{20}Hz[/tex]
Re-arranging the equation, we find its wavelength:
[tex]\lambda=\frac{c}{f}=\frac{3\cdot 10^8}{7.16\cdot 10^{20}}=4.19\cdot 10^{-13} m[/tex]
c)
The size of the nuclear radius is approximately
[tex]d \sim 10^{-15} m[/tex]
While we see that the wavelength of this photon is
[tex]\lambda=4.19\cdot 10^{-13} m[/tex]
Therefore, the ratio between the wavelength of the photon and the nuclear radius is
[tex]\frac{\lambda}{d}=\frac{\sim 10^{-13}}{\sim \cdot 10^{-15}}=100[/tex]
So, the wavelength of the photon is approximately a factor 100 times larger than the nuclear radius.
Learn more about photons:
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