To solve, apply the Planck's equation.
`E = hv`
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)`
This is the frequency of the gamma ray.
To determine its wavelength, apply the wave equation.
`c= lambda v`
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.