The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequency of an emitted gamma particle having the energy of 1.29 × 1011 j/mol. Enter your answer in scientific notation.

Respuesta :

1. Frequency: [tex]3.23\cdot 10^{20} Hz[/tex]

The energy given is the energy per mole of particles:

[tex]E=1.29\cdot 10^{11} J/mol[/tex]

1 mole contains a number of Avogadro of particles, [tex]N_A[/tex], equal to

[tex]N_A=6.022\cdot 10^{23}[/tex] particles

So, by setting the following proportion, we can calculate the energy of a single photon:

[tex]1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J[/tex]

This is the energy of a single photon; now we can calculate its frequency by using the formula:

[tex]E_1 = hf[/tex]

where

[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck's constant

f is the photon frequency

Solving for f, we find

[tex]f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz[/tex]

2. Wavelength: [tex]9.29\cdot 10^{-13} m[/tex]

The wavelength of the photon is given by the equation:

[tex]\lambda=\frac{c}{f}[/tex]

where

[tex]c=3\cdot 10^8 m/s[/tex]

is the speed of the photon (the speed of light). Substituting,

[tex]\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m[/tex]