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 multiplied by 10-19 J). What is the frequency (in Hz) and the wavelength (in m) of this gamma ray?

Convert 1.33 MeV to J, then

E = hc/wavelength to find wavelength and
c = frequency*wavelength to find frequency.

To find the frequency and wavelength of the gamma ray emitted by Cobalt-60, we can use the equation relating energy, frequency, and wavelength.

The equation is:
E = h * f

Where:
E is the energy of the gamma ray (in Joules)
h is Planck's constant (6.626 × 10^-34 J·s)
f is the frequency of the gamma ray (in Hz)

First, let's convert the energy from MeV to Joules:
1 MeV = 1.602 × 10^-13 J

Given that the energy of the gamma ray is 1.33 MeV, we can convert it to Joules:
E = 1.33 MeV * (1.602 × 10^-13 J/MeV)
E ≈ 2.132 × 10^-13 J

Now we can rearrange the equation to solve for the frequency:
f = E / h

Substituting the values:
f = 2.132 × 10^-13 J / (6.626 × 10^-34 J·s)

Now we can calculate the frequency (f) of the gamma ray by dividing the energy (E) by Planck's constant (h).

f ≈ 3.22 × 10^20 Hz

To find the wavelength (λ), we can use the equation:
c = λ * f

Where:
c is the speed of light (approximately 3 × 10^8 m/s)
λ is the wavelength of the gamma ray (in meters)

Rearranging the equation to solve for the wavelength:
λ = c / f

Plugging in the values:
λ = (3 × 10^8 m/s) / (3.22 × 10^20 Hz)

λ ≈ 9.32 × 10^-13 m

Therefore, the frequency of the gamma ray emitted by Cobalt-60 is approximately 3.22 × 10^20 Hz, and the wavelength is approximately 9.32 × 10^-13 m.