Question

Cobaltâ’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 4.70 MeV (million electron volts; 1 eV = 1.602 Ă— 10â’19 J). What is the frequency (in Hz) and the wavelength (in m) of this gamma ray? Enter your answers in scientific notation.

Answer 1

Energy of gamma rays is given by equation

`E = h\\u`

here we know that

h = Planck's constant

`\\u = frequency`

now energy is given as

`E = 4.70 MeV = 4.70 * 10^6 * 1.6 * 10^(-19)`

`E = 7.52 * 10^(-13) J`

now by above equation

`E = h\\u`

`7.52 * 10^(-13) = 6.6 * 10^(-34) \\u`

`\\u = 1.14 * 10^(21) Hz`

now for wavelength we can say

`\lambda = (c)/(\\u)`

`\lambda = (3* 10^8)/(1.14 * 10^(21))`

`\lambda = 2.63 * 10^(-13) m`