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Express the frequency in inverse seconds. n=4-->n=3. Can you please tell me what formulas to use, because I have like 6 more of these to do and I can't figure it out. Thanks.​

Express the frequency in inverse seconds n4gtn3 Can you please tell me what formulas to use because I have like 6 more of these to do and I cant figure it out T class=

Respuesta :

Edufirst

Answer:

  • 1.60×10¹⁴ s⁻¹

Explanation:

When an electron jumps from one energy level to a lower energy level some energy is released in the form of a photon.

The difference in energy between the two levels is the energy of the photon and that energy is related to the frequency of the photon by the Einstein - Planck equation:

  • E = hν

Where,

  • E = energy of the photon,
  • h = 6.626×10⁻³⁴ J.s, Planck constant, and
  • ν = frequency of the photon.

So, to find the frequency you must first find the energy.

The transition energy can be calculated using the formula:

  • Eₙ = - E₀ ( 1/ n²)

Where E₀ = 13.6 eV ( 1 eV = 1.602×10⁻¹⁹ Joules) and n = 1,2,3,...

So, the transition energy between n = 4 and n = 3 will be:

  • ΔE = - E₀ [ 1/4² - 1/3²] = - 13.6 eV [1/16 - 1/9] = 0.6611. . .eV

  • ΔE = 1.602×10⁻¹⁹ Joules/eV  × 0.6611... eV = 1.0591 ×10⁻¹⁹ Joules

Now you can use the Einstein - Planck equation:

  • E = hν
  • ν = E / h
  • ν = 1.0591 ×10⁻¹⁹ J / 6.626×10⁻³⁴ J.s  = 1.60×10¹⁴ s⁻¹ (rounded to 3 significant figures).
BladeRunner212

v = 1.597 x 10¹⁴ s⁻¹

Further explanation

Given:

  • n₁ = 4 as the initial stationary orbit
  • n₂ = 3 as the final stationary orbit

Question:

Express the frequency in inverse seconds.

The Process:

To solve this problem, we will use the Planck-Einstein formula, which is:

[tex]\boxed{ \ E_1 - E_2 = hv \ }[/tex]

with,

  • E₁ = The energy of the photon in the initial stationary orbit (n₁).
  • E₂ = The energy of the photon in the final stationary orbit (n₂).
  • h = Planck's constant = 6.626 x 10⁻³⁴ J.s
  • v = frequency (Hz or s⁻¹)

        where v represents the frequency of the emitted (radiated) photon.

And remember the ionization energy based on Bohr's theory, namely:

[tex]\boxed{ \ \Delta E = E_1 - E_2 = -13.6 \Big( \frac{1}{n_1^2} - \frac{1}{n_2^2} \Big) \ }[/tex] in eV (electronvolt).

[tex]\boxed{ \ 1 \ eV = 1.6 \times 10^{-19} \ joule \ }[/tex]

Step-1

[tex]\boxed{ \ E_1 - E_2 = -13.6 \Big( \frac{1}{4^2} - \frac{1}{3^2} \Big) \ }[/tex]

[tex]\boxed{ \ E_1 - E_2 = -13.6 \Big( \frac{1}{16} - \frac{1}{9} \Big) \ }[/tex]

[tex]\boxed{ \ E_1 - E_2 = -13.6 \Big( \frac{9 - 16}{144} \Big) \ }[/tex]

[tex]\boxed{ \ E_1 - E_2 = -13.6 \Big( - \frac{7}{144} \Big) \ }[/tex]

We got ΔE = 0.661 eV.

Let us convert eV to joules.

[tex]\boxed{ \ \Delta E = 0.661 \ eV \times 1.6 \times 10^{-19} \ \frac{J}{eV} \ }[/tex]

Thus, the energy emitted by photons from n = 4 to n = 3 is ΔE = 1.058 x 10⁻¹⁹ J.

Step-2

Now let us calculate the frequency of photons emitted during the transition.

[tex]\boxed{ \ hv = \Delta E \ } \rightarrow \boxed{ \ v = \frac{\Delta E}{h} \ }[/tex]

[tex]\boxed{ \ v = \frac{1.058 \times 10^{-19}}{6.626 \times 10^{-34}} \ }[/tex]

Thus, the frequency of photons emitted during the transition is [tex]\boxed{ \ v = 1.597 \times 10^{14} \ Hz \ }[/tex] or  [tex]\boxed{ \ v = 1.597 \times 10^{14} \ s^{-1} \ }[/tex]

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