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Step-by-Step Solution
Step 1: Write the formula for the wavelengths of the Balmer series
The Balmer series corresponds to electron transitions ending at the energy level n=2 in the hydrogen atom. The wavelength \lambda of a spectral line for hydrogen (with atomic number Z=1 ) is given by:
\displaystyle \frac{1}{\lambda} \;=\; R \, Z^2 \left(\frac{1}{n_{1}^2} \;-\; \frac{1}{n_{2}^2}\right)
Here, R is the Rydberg constant, n_1 is the lower energy level (2 for Balmer series), and n_2 is the upper energy level.
Step 2: Identify the transitions for the first two Balmer lines
The first member of the Balmer series is the transition from n=3 to n=2 .
The second member of the Balmer series is the transition from n=4 to n=2 .
Step 3: Express the given information for the first Balmer line
The first member of the Balmer series (from n=3 to n=2 ) has wavelength \lambda_1 = 6561 \text{ Å} . This can be written as:
\displaystyle \frac{1}{\lambda_1} \;=\; R \left(\frac{1}{2^2} - \frac{1}{3^2}\right)
\;=\; R \left(\frac{1}{4} - \frac{1}{9}\right)
\;=\; R \left(\frac{9 - 4}{36}\right)
\;=\; \frac{5}{36} R \,.
Step 4: Write the expression for the second Balmer line
The second member of the Balmer series (from n=4 to n=2 ) has wavelength \lambda_2 , given by:
\displaystyle \frac{1}{\lambda_2} \;=\; R \left(\frac{1}{2^2} - \frac{1}{4^2}\right)
\;=\; R \left(\frac{1}{4} - \frac{1}{16}\right)
\;=\; R \left(\frac{4 - 1}{16}\right)
\;=\; \frac{3}{16} R \,.
Step 5: Form the ratio of second to first wavelength
Taking the ratio \frac{\lambda_2}{\lambda_1} using their reciprocal forms:
\displaystyle
\frac{\lambda_2}{\lambda_1}
= \frac{\frac{1}{\left(\frac{3}{16} R\right)}}{\frac{1}{\left(\frac{5}{36} R\right)}}
= \frac{\frac{16}{3 R}}{\frac{36}{5 R}}
= \frac{16}{3} \times \frac{5}{36}
= \frac{80}{108}
= \frac{20}{27}
\,.
Step 6: Calculate \lambda_2 from \lambda_1
Since \lambda_1 = 6561 \text{ Å} , we have:
\displaystyle
\lambda_2
= \frac{20}{27} \times 6561 \text{ Å}
= 4860 \text{ Å}
= 486 \text{ nm}
\,.
Final Answer
The wavelength of the second member of the Balmer series is 486 \text{ nm} .
