15 Aug, 2024
· Chemistry

What's the rearranged Rydberg(Bohr) equation applied to find the ni or nf

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Long Explanation

Explanation

Rearranged Rydberg-Bohr Equation to Find ni or nf

The Rydberg-Bohr equation is applicable for calculating the wavelengths of spectral lines in hydrogen-like atoms. To find the initial (nin_i) or final (nfn_f) principal quantum numbers, the equation can be rearranged. The general form of the Rydberg formula for the wavelength λ\lambda is:

1λ=R(1nf2−1ni2)\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

where RR is the Rydberg constant. To isolate nin_i or nfn_f, we follow these steps:

  1. Isolating 1ni2\frac{1}{n_i^2} or 1nf2\frac{1}{n_f^2}:

1λR=1nf2−1ni2\frac{1}{\lambda R} = \frac{1}{n_f^2} - \frac{1}{n_i^2}

Now, rearrange it to solve for the desired term:

1ni2=1nf2−1λR\frac{1}{n_i^2} = \frac{1}{n_f^2} - \frac{1}{\lambda R}

or

1nf2=1ni2+1λR\frac{1}{n_f^2} = \frac{1}{n_i^2} + \frac{1}{\lambda R}

  1. Solving for nin_i:

To find nin_i, use the isolated term for 1ni2\frac{1}{n_i^2}:

1ni2=1nf2−1λR\frac{1}{n_i^2} = \frac{1}{n_f^2} - \frac{1}{\lambda R}

Then take the reciprocal to find ni2n_i^2:

ni2=1(1nf2−1λR)n_i^2 = \frac{1}{\left( \frac{1}{n_f^2} - \frac{1}{\lambda R} \right)}

Finally, take the square root to find nin_i:

ni=1(1nf2−1λR)n_i = \sqrt{\frac{1}{\left( \frac{1}{n_f^2} - \frac{1}{\lambda R} \right)}}

  1. Solving for nfn_f:

To find nfn_f, use the isolated term for 1nf2\frac{1}{n_f^2}:

1nf2=1ni2+1λR\frac{1}{n_f^2} = \frac{1}{n_i^2} + \frac{1}{\lambda R}

Then take the reciprocal to find nf2n_f^2:

nf2=1(1ni2+1λR)n_f^2 = \frac{1}{\left( \frac{1}{n_i^2} + \frac{1}{\lambda R} \right)}

Finally, take the square root to find nfn_f:

nf=1(1ni2+1λR)n_f = \sqrt{\frac{1}{\left( \frac{1}{n_i^2} + \frac{1}{\lambda R} \right)}}

  1. Summary

  • For nin_i:
ni=1(1nf2−1λR)n_i = \sqrt{\frac{1}{\left( \frac{1}{n_f^2} - \frac{1}{\lambda R} \right)}}
  • For nfn_f:
nf=1(1ni2+1λR)n_f = \sqrt{\frac{1}{\left( \frac{1}{n_i^2} + \frac{1}{\lambda R} \right)}}

Use these rearranged equations to determine the initial or final principal quantum numbers in atomic transitions.

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Johnathan Clark

Chemistry Content Writer at Math AI

Johnathan Clark, with a Master's in Chemistry from the University of São Paulo, is a young high school chemistry teacher and part-time contract writer. His engaging classroom experiments translate into compelling written content that makes chemistry exciting and practical.

chemistry
Concept

1. Rydberg Constant And Its Role In Spectral Lines Calculation

Explanation

The Rydberg constant is a fundamental physical constant that plays a crucial role in the calculation of spectral lines in atomic physics. It is denoted by R∞R_{\infty} and has the value

R∞≈1.097×107 m−1.R_{\infty} \approx 1.097 \times 10^7 \, \text{m}^{-1}.

Formula for Spectral Lines

The Rydberg constant is prominently used in the Rydberg formula, which is used to predict the wavelengths (λ\lambda) of the photons emitted or absorbed by electrons transitioning between energy levels in a hydrogen atom. The formula is given by:

1λ=R∞(1n12−1n22)\frac{1}{\lambda} = R_{\infty} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

where:

  • n1n_1 and n2n_2 are integers representing the principal quantum numbers of the electron's initial and final energy levels, with n2>n1n_2 > n_1.

Energy Levels and Transitions

The relation between the Rydberg constant and the energy levels of an electron in a hydrogen atom is fundamental to understanding spectral lines. The energy of an electron in a hydrogen atom is given by:

En=−hcR∞n2E_n = - \frac{hcR_{\infty}}{n^2}

Here:

  • hh is Planck's constant,
  • cc is the speed of light,
  • nn is the principal quantum number (e.g., n=1,2,3,…n = 1, 2, 3, \ldots).

When an electron transitions between levels n2n_2 and n1n_1, it emits or absorbs a photon with energy corresponding to the difference in these energy levels.

Importance of the Rydberg Constant

Key Points:

  • The Rydberg constant provides a way to predict the spectral lines of hydrogen, which can be extended to other elements through more complex formulations.
  • It is critical to quantum mechanics and atomic physics, helping to determine the wavelengths of emitted or absorbed light.
  • Accurate knowledge of R∞R_{\infty} allows for precise calculations of these spectral lines, aiding in various scientific investigations, including the study of other elements and astrophysical phenomena.

In summary, the Rydberg constant is essential for understanding atomic spectra and the quantized nature of energy levels in atoms, providing a cornerstone for both theoretical and applied physics.

Concept

2. Rearrangement Of Equations To Isolate Variables

Explanation

When working with equations, it is often necessary to isolate a specific variable. This process involves manipulating the equation so that the variable of interest is alone on one side of the equation.

Basic Steps

  1. Identify the variable you want to isolate.
  2. Use inverse operations to move other terms to the opposite side of the equation. Inverse operations include:
    • Addition and subtraction
    • Multiplication and division
    • Exponentiation and root extraction

Example

Consider the equation:

3x+2y=123x + 2y = 12

If we want to isolate xx, we follow these steps:

  1. Subtract 2y2y from both sides:

    3x+2y−2y=12−2y3x + 2y - 2y = 12 - 2y

    Simplifying, we get:

    3x=12−2y3x = 12 - 2y

  2. Divide both sides by 3:

    3x3=12−2y3\frac{3x}{3} = \frac{12 - 2y}{3}

    Simplifying, we get:

    x=12−2y3x = \frac{12 - 2y}{3}

Complex Equations

For more complex equations involving quadratic terms, fractions, or multiple variables, the process can be more involved but follows the same principles.

Example with Fractions

Let's isolate yy in:

2y−35=x+4\frac{2y - 3}{5} = x + 4

  1. Multiply both sides by 5 to eliminate the denominator:

    2y−3=5(x+4)2y - 3 = 5(x + 4)

    Simplifying, we get:

    2y−3=5x+202y - 3 = 5x + 20

  2. Add 3 to both sides:

    2y−3+3=5x+20+32y - 3 + 3 = 5x + 20 + 3

    Simplifying, we get:

    2y=5x+232y = 5x + 23

  3. Divide by 2:

    2y2=5x+232\frac{2y}{2} = \frac{5x + 23}{2}

    Simplifying, we get:

    y=5x+232y = \frac{5x + 23}{2}

Summary

The key steps to isolating a variable are:

  • Use inverse operations to move terms across the equation.
  • Perform algebraic simplification as you go.
  • Ensure you manipulate both sides of the equation equally to maintain equality.

By following these steps, you can effectively isolate any variable in an equation, simplifying problem-solving in various mathematical contexts.