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Cobalt-60 is a radioactive isotope used to treat cancers. A gamma ray emitted by this isotope has an energy of 1.33 MeV (million electron volts; 1 eV = 1.602 x 10¹⁹ J). What is the frequency (in Hz) and the wavelength (in m) of this gamma ray?

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Answer:

E = 1.33 MeV = 2.13 x [tex]10^{-13}[/tex] J

v = wavelength = E / h = 2.13 x [tex]10^{-13}[/tex] / 6.626 x [tex]10^{-34}[/tex] = 3.2 x [tex]10^{20}[/tex] m

f = frequency = c / 3.2 x [tex]10^{20}[/tex] m = 3 x [tex]10^{8}[/tex] / 3.2 x [tex]10^{20}[/tex] = 9.375 x [tex]10^{-13}[/tex] Hz

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