The first ionization potential for calcium (Z = 20, A = 40) is 6.11 eV. Singly-ionized calcium (Ca+) produces two very strong absorption lines in the Sun’s spectrum discovered by Joseph Fraunhofer in 1814, who named them "H" and "K" (he didn’t know they were from calcium, as this was >100 years before the development of quantum mechanics). Both lines always appear together, with lambda subscript H equals 3968 end subscript Å and lambda subscript K equals 3933 Å; hence they are called a "doublet A. What is the speed of an electron that has just barely enough kinetic energy to collisionally ionize a neutral calcium atom? What is the speed of a calcium ion with this same kinetic energy?
B. What is the temperature T of a gas in which the average particle energy is just barely sufficient to ionize a neutral calcium atom?
C. The lower energy level of both lines is the ground state of Cat. What is the difference in energy in eV) between the two states that correspond to the upper energy levels of the Hand Klines, respectively? How does this compare to the energy of a calcium K photon? Can these two lines can be formed by transitions to upper energy levels with different principal quantum numbers (different n), or do they represent transitions with the same n but some different higher-order quantum number? Explain your reasoning based on your understanding of the general behavior of atomic energy levels (En).

A) v = 1.47 10⁶ m / s, v = 0.5426 10⁴ m / s , B)  T = 4.7 10⁴ K, C) n₂ = 42

Explanation:

A)  For this part, let's calculate the speed of an electron that has an energy of 6.11 eV.

Let's reduce the units to the SI system

        E₀ = 6.11eV (1.6 10⁻¹⁹ J / 1eV) = 9.776 10⁻¹⁹ J

The kinetic energy of the electron is

        K = ½ m v²

         E₀ = K

         v = √ 2E₀ / m

         v = √ (2 9.776 10⁻¹⁹ / 9.1 10⁻³¹)

         v = √ (2.14857 10¹²)

         v = 1.47 10⁶ m / s

now the speed of a calcium ion is asked, let's find sum

        m = 40 1.66 10⁻²⁷ = 66.4 10⁻²⁷ kg

        v = √ (2E₀ / M)

         v = √ (2 9.776 10⁻¹⁹ / 66.4 10⁻²⁷)

        v = √ (0.2994457 10⁸)

        v = 0.5426 10⁴ m / s

B) the terminal energy of an ideal gas is

             E = 3/2 kT

              T = ⅔ E / k

              T = ⅔ (9,776 10-19 / 1,381 10-23)

              T = 4.7 10⁴ K

C) To calculate the energy of these lines we use the Planck expression

              E = h f

where wavelength and frequency are related

              c =λ f

              f = c /λ

let's substitute

              E = h c /λ

let's look for the energies

λ = 396.8 nm

  E₁ = 6.63 10⁻³⁴ 3 10⁸ / 396.8 10⁻⁹

           E₁ = 5.0126 10⁻¹⁹ J

λ = 393.3 nm

           E₂ = 6.63 10⁻³⁴ 3 10⁸ / 3.93.3 10⁻⁹

           E₂ = 5.0572 10⁻¹⁹ J

The difference in energy between these two states is

          ΔE = E₂ -E₁

          ΔE = (5.0572 - 5.0126) 10⁻¹⁹ J

          ΔE = 0.0446 10⁻¹⁹ J

let's reduce eV

         ΔE = 0.0446 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

         ΔE = 2.787 10⁻² eV

Now let's use Bohr's atomic model for atoms with one electron,

               E = -13.606 Z² / n²

where 13,606 eV is the energy of the base state of the Hydrogen atom, Z is the atomic number of Calcium

               n = √ (13.606 Z² / E)

λ = 396.8 nm

E₁ = 5.0126 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹J) = 3.132875 eV

               n₁ = √ (13.606 20² / 3.132875)

               n₁ = 41.7

since n must be an integer we take

               n₁ = 42

λ = 393.3 nm

E₂ = 5.0572 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹ J) = 3.16075 eV

              n₂ = √ (13.606 20² / 3.16075)

              n₂ = 41.5

Again we take n as an integer

               n₂ = 42

We can see that the two lines have the same principal quantum number, so for the difference of these energies there must be other quantum numbers, which are not in the Bohr model, because of the small difference they are possibly due to small numbers of the moment angular orbital or spin

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