Question

An electron from a Ti ^ + 2 hydrogen ion leaps from one orbit with radius 13.25 angstrom to another orbit with radius 2.12 angstrom. determine the energy (Joule) e produced in said transition and the wavelength (in cm)

Answer 1

ΔE = 59.75 A,

Step-by-step 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