Askiitians Tutor Team
Last Activity: 23 Days ago
To determine the value of 'n' corresponding to the observed wavelength of 4341 Å in the Balmer series, we can use the Rydberg formula. The Balmer series describes the spectral lines of hydrogen when an electron transitions from a higher energy level (n) to the second energy level (n=2). The Rydberg formula for the wavelengths of these transitions is given by:
Rydberg Formula
The formula is expressed as:
1/λ = R_H (1/2² - 1/n²)
Where:
- λ is the wavelength of the emitted light.
- R_H is the Rydberg constant, approximately 1.097 × 107 m-1.
- n is the principal quantum number of the higher energy level (n > 2).
Step-by-Step Calculation
First, we need to convert the wavelength from angstroms to meters for consistency with the Rydberg constant:
4341 Å = 4341 × 10-10 m = 4.341 × 10-7 m.
Next, we can rearrange the Rydberg formula to solve for 'n':
1/n² = 1/2² - 1/(R_H * λ)
Now, substituting the known values:
1/n² = 1/4 - 1/(1.097 × 107 m-1 * 4.341 × 10-7 m)
Calculating the second term:
1/(1.097 × 107 * 4.341 × 10-7) = 1/(4.77) ≈ 0.209.
Now, substituting this back into the equation:
1/n² = 0.25 - 0.209 = 0.041.
Taking the reciprocal gives:
n² = 1/0.041 ≈ 24.39.
Finally, taking the square root gives:
n ≈ 4.94.
Determining the Principal Quantum Number
Since 'n' must be a whole number, we round 4.94 to the nearest integer, which is 5. Therefore, the value of 'n' corresponding to the observed wavelength of 4341 Å in the Balmer series is:
n = 5
Summary
In summary, the observed wavelength of 4341 Å corresponds to a transition from the fifth energy level (n=5) to the second energy level (n=2) in the hydrogen atom's Balmer series. This process illustrates how energy levels in atoms are quantized and how they emit specific wavelengths of light when electrons transition between these levels.