Answer:
ΔE = 59.75 A,
Explanation:
Titanium has 3 electrons in its last shell, as it is doubly ionized, it is left with a single electron in this shell, which is why it behaves like a hydrogen-type atom, consequently we can use Bohr's atomic theory
rₙ = a₀ /Z n²
Eₙ = 1k e² / 2a₀ (Z² / n²)
Where a₀ is Bohrd's atomic radius so = 0.529 núm
Let's find out what quantum number n has each orbit
rn = 13.25 A = 1.325 nm
for Titanium with atomic number 22
n² = Z rₙ / a₀
n = √ (22 (1.325 / 0.529))
n = 7.4
since N is an entry we take
n = 7
rn = 2.12 A = 0.212 nm
n = √ (22 / 0.529) 0.212
n = 3
With these values we can calculate the energy of the transition from level ne = 7 to level no = 3
ΔE = ka e2 Z2 / 2ao (1n02 - 1 / nf2)
ΔE = 9 10⁹ 1.6 10⁻¹⁹ 22² (2 0.529 10⁻⁹) (1/3² - 1/7²)
ΔED = 6.5875 10² (0.111 - 0.0204)
ΔE = 59.75 A
let us be the Planck relation between energy and frequency
E = h f
the frequency is related to the speed of light
c = λ f
f = c / λ
we substitute
E = h c /y
E = ΔE
h c /λ = E
λ = 6.63 10-34 3 108 / 59.75
λ= 3.01939 10⁻²⁴ m
λ = 3.01939 10⁻²² cm
