Question

You have a great summer job working in a cancer research laboratory. Your team is trying to construct a gas laser that will give off light of an energy that will pass through the skin but be absorbed by cancer tissue. You know that an atom emits a photon (light) when an electron goes from a higher energy orbit to a lower energy orbit. Only certain orbits are allowed in a particular atom. To begin the process, you calculate the energy of photons emitted by a Helium ion in which the electron changes from an orbit with a radius of 0.30 nanometers to another orbit with a radius of 0.20 nanometers. A nanometer is 10-9m. The helium nucleus consists of two protons and two neutrons. (In reality, the energy levels of electrons in atoms are governed by quantum mechanics, but this classical calculation is a good first estimate.)

Answer 1

Answer:

ΔE = 7.559 eV ,     λ = 1,645 10⁻⁷ m

Explanation:

For this exercise we can use the Bohr model for ionized atom with only one free electron,

         r_n = n² a₀ / Z

         E_n = -13,606 Z² / n²

Where a₀ is the Bohr radius of the hydrogen atom (a₀ = 0.0529 nm), Z is the atomic number of the atom under study and 13.606 eV is the energy of the ground state of Hydrogen.

In our case the Helium atom has two protons Z = 2

let's calculate the quantum number and the energy of each orbit

r_n = 0.30 nm

          n₁ = √ (r_n Z / a₀)

          n₁ = √ (0.30 2 / 0.0529)

           

Note that we do not have to reduce the radius since they are all in nanometers

          n₁ = 3.3

since n is an integer we approximate it to

         n₁ = 3

r_n = 0.20 nm

          n₂ = √ (0.2 2 / 0.0529)

          n₂ = 2.7

To approximate this value we must assume that there could be some error in the medicinal radio,

          n₂ = 2

having the quantum numbers of the two radius we can calculate their energy

        E₃ = - 13,606 2²/3²

        E₃ = - 6.047 eV

   

        E₂ = -13.606 2²/2²

         E₂ = -13.606 eV

the energy of the emitted photon is

          ΔE = E₃ - E₂

          ΔE = -6.047 + 13.606

          ΔE = 7.559 eV

You do not indicate in the exercise if you want the energy or the wavelength of the photon,

         

to find the wavelength We use the Planck relation

          E = h f

          c = λ f

          E = h c /λ

          λ = h c / E

we must reduce the energy to the SI system

          E = 7.559 ev (1.6 10⁻¹⁹ J / 1eV) = 12.09 10⁻¹⁹ J

         

          λ = 6.63 10⁻³⁴ 3 10⁸ / 12.09 10⁻¹⁹

          λ = 1,645 10⁻⁷ m