Question

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

Answer 1

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

$E = -13.6$×$Z$²/$n$²

where,

$E$ = energy of  orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

$E$₁ = -13.6× 1²/1² = -13.6eV

Energy of n = 2 in an hydrogen atom:

$E$₂ = -13.6 × 1²/2² = -3.40eV

Energy of n = 3 in an hydrogen atom:

$E$₃ = -13.6× 1²/3² = -1.51eV

Energy of n = 4 in an hydrogen atom:

$E$₄ = -13.6× 1²/4² = -0.85eV

Energy of n = 5 in an hydrogen atom:

$E$₃ = -13.6× 1²/ 5² = -0.54eV

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

Δ$E$₁ = 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 :

$E$₄< 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:

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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

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