# 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.