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Simora [160]
3 years ago
10

27/13Al+4/2He >>>> 30/15P+

Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
5 0

Answer

Explanation:

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

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3 years ago
Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
ololo11 [35]

Hence the order of the transition in order of increasing frequency of the photon absorbed or emitted  will be : d < a < c < b

E = -13.6×Z²/n²

where,

E = energy of  orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

E₁ = -13.6× 1²/1² = -13.6eV

Energy of n = 2 in an hydrogen atom:

E₂ = -13.6 × 1²/2² = -3.40eV

Energy of n = 3 in an hydrogen atom:

E₃ = -13.6× 1²/3² = -1.51eV

Energy of n = 4 in an hydrogen atom:

E₄ = -13.6× 1²/4² = -0.85eV

Energy of n = 5 in an hydrogen atom:

E₃ = -13.6× 1²/ 5² = -0.54eV

a) n = 2 to n = 4 (absorption)

ΔE₁ = E₄ - E₂ = -0.85- (- 3.40) = 2.55eV

b) n = 2 to n = 1 (emission)

Δ E₂ = E₁ - E₂ = -13.6 - (- 3.40) = -10.2eV

Negative sign indicates that emission will take place.

c) n = 2 to n = 5 (absorption)

ΔE₃ = E₅ - E₂ = -0.54 -( -3.40) = -2.85eV

d) n = 4 to n = 3 (emission)

ΔE₄ = E₃ - E₄ = -1.51 - (- 0.85) = - 0.66eV

Negative sign indicates that emission will take place.

According to Planck's equation, higher the frequency of the wave higher will be the energy:

E = hv

h = Planck's constant

v =frequency of the wave

So, the increasing order of magnitude of the energy difference :

E₄< E₁ <E₃ <E₂

The  H atom electron transitions in order of increasing frequency of the photon absorbed or emitted will be d < a < c < b

: d < a < c < b

To learn more about transitions visit the link:

brainly.com/question/28304182?referrer=searchResults

The question is incomplete , complete question is:

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

(a) n = 2 to n = 4

(b) n = 2 to n = 1

(c) n = 2 to n = 5

(d) n = 4 to n = 3

#SPJ4

6 0
1 year ago
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