Question

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

Answer 1

The order of frequency is d < a < c < b

$E_{n} = 13.6 * z^{2} / n^{2}$ eV

where z = atomic mass number

n = energy level

For hydrogen z = 1

Therefore, Energy for n = 1

$E_{n} = 13.6 * 1^{2} / 1^{2}$ eV

     = -13.6 eV

for n = 2

$E_{n} = 13.6 * 1^{2} / 2^{2}$ eV

     = -3.40 eV

for n = 3

 $E_{n} = 13.6 * 1^{2} / 3^{2}$ eV

      = -1.51 eV

for n = 4

 $E_{n} = 13.6 * 1^{2} / 4^{2}$ eV

      = -0.85 eV

for n = 5

 $E_{n} = 13.6 * 1^{2} / 5^{2}$ eV

      = -0.544 eV

n = 2 to n = 4 (absorption)

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

n = 2 to n = 1 (emission)

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

The negative sign indicates that emission will take place.

n = 2 to n = 5 (absorption)

ΔE = E5 - E2 = -0.544 - (-3.40) = 2.856 eV

n = 4 to n = 3 (emission)

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

We know that

E = h * υ

Therefore, Energy is proportional to frequency.

So increasing the order of energy is

E4  < E1  < E3  <  E2

order of frequency is

d < a < c < b

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brainly.com/question/17058029

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