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yuradex [85]
3 years ago
11

Please help me figure this out! God bless you <3

Chemistry
1 answer:
statuscvo [17]3 years ago
3 0
South ????????????????????????
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what electronic transition in a hydrogen atom starting from n=7 energy level will produce infrared radiation with energy of 55.1
Lostsunrise [7]

Answer:

(<em>n</em> = 7) ⟶ (<em>n</em> = 4)

Explanation:

1. Convert the energy to <em>joules per mole of electrons</em>.  

<em>E</em> = 55.1 × 1000 = 55 100 J/mol

2. Convert the energy to <em>joules per electron </em>

<em>E</em> = 55 100/(6.022 × 10²³)

<em>E</em> = 9.150 × 10⁻²⁰ J/electron

3. Use the Rydberg equation to <em>calculate the transition </em>

Rydberg's original formula was in terms of wavelengths, but we can rewrite it to have the units of energy. The formula then becomes  

\Delta E = R_{\text{H}} (\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ n_{i}^{2}})

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R_{\text{H}} = the Rydberg constant = 2.178 × 10⁻¹⁸ J

n_{i} and n_{f} are the initial and final energy levels.

9.150 \times 10^{-20} = 2.178 \times 10^{-18}(\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 7^{2}})      

\frac{9.150 \times 10^{-20} }{2.178 \times 10^{-18}} = \frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 49}

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01} + \frac{1 }{49 }

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}

\frac{ 1}{ n_{f}^{2}} = \text{0.062 42}

n_{f}^{2} = \frac{1 }{ \text{0.062 42}}

n_{f}^{2} = 16.02

n_{f} = \sqrt{16.02}

n_{f} = 4.003 \approx 4

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Wat substance cannot be separated physically or chemically?
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