Physics: Cobalt-60 is an artificial radioisotope that is produced in a nuclear reactor and is used as a gamma-ray source in the treatment of certain types of cancer. if the wavelength of the gamma radiation from a cobalt-60 source is 1.00 × 10−3 nm, calculate the energy of a photon of this radiation.

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Cobalt-60 is an artificial radioisotope that is

ptonygonzalez701

09.01.2022

$1.9878\times 10^{-13}\ Joules$

Explanation:

c = Speed of light = $3\times 10^8\ m/s$

$\lambda$ = Wavelength = $1\times 10^{-3}\times 10^{-9}\ m$

h = Planck's Constant = $6.626\times 10^{-34}\ m^2kg/s$

Energy

$E=h\nu$

Frequency

$\nu=\frac{c}{\lambda}\\\Rightarrow \nu=\frac{3\times 10^8}{1\times 10^{-3}\times 10^{-9}}\\\Rightarrow \nu=3\times 10^{20}\ Hz$

$E=h\nu\\\Rightarrow E=6.626\times 10^{-34}\times 3\times 10^{20}\\\Rightarrow E=1.9878\times 10^{-13}\ J$

The energy of a photon is $1.9878\times 10^{-13}\ Joules$

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