Cobalt-60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this isotope has an energy of 9.70 MeV (million electron volts; 1 eV=1.602x10^-19 J). What is the frequency (in Hz) and the wavelength (in m) of this gamma ray? Answers must be in scientific notation.

To solve, apply the Planck's equation.
E = hv
where 
E is the energy  of photons in Joules
h is the Planck's constant 6.63xx 10^(-34) J*s , and
v is the frequency in Hertz.
Plugging in the known values, the formula becomes:
9.70MeV = (6.63xx10^(-34)J*s)v
9.70xx10^6eV = (6.63xx10^(-34)J*s)v
In order for the unit to be consistent, convert the given energy to Joules.
9.7xx10^6 eV * (1.602xx10^(-19)J)/(1eV) = (6.63xx10^(-34) J*s)v
1.55394 xx10^(-12)J = (6.63xx10^(-34)J*s)v
Then, isolate the v .
v = (1.55394xx10^(-12)J)/(6.63xx10^(-34)J*s)
v=(2.34xx10^(21) )/s
v=2.34xx10^(21) Hz
This is the frequency of the gamma ray.
To determine its wavelength, apply the wave equation.
c= lambda v
where
c is the speed of light, 3xx10^8 m//s and
lambda is the wavelength in meters.
Plugging in the value of c and v, the formula becomes:
3xx10^8 m//s = lambda( 2.34xx10^21Hz)
And, isolate the wavelength.
lambda = (3xx10^8 m//s)/(2.34xx10^21 Hz)
lambda = 1.28 xx10^(-13) m
 
Therefore, the frequency and wavelength of the gamma ray is 2.34xx10^21 Hz and 1.28xx10^(-13)m , respectively.