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Chapter 18: Problem 2539

In terms of Rydberg constant \(R\). The wave number of first Balmer line is (A) \((5 \mathrm{R} / 36)\) (B) \((8 \mathrm{R} / 9)\) (C) \(\mathrm{R}\) (D) \((8 \mathrm{R} / 20)\)

Short Answer

Expert verified
The wave number of the first Balmer line in terms of the Rydberg constant R can be calculated using the Rydberg formula for wave number \(\bar\nu = R \times ( \frac{1}{n_1^2} - \frac{1}{n_2^2})\). For the first Balmer line, \(n_1 = 2\) and \(n_2 = 3\). Plugging these values and simplifying the expression, we get \(\bar\nu = \frac{5R}{36}\), which corresponds to option (A).

Step by step solution

01

Recall the Rydberg formula for wave number

The Rydberg formula for wave number is given by: \(\bar\nu = R \times ( \frac{1}{n_1^2} - \frac{1}{n_2^2})\) Here, \(\bar\nu\) is the wave number, \(R\) is the Rydberg constant, \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.
02

Apply the formula for the first Balmer line

For the first Balmer line, we have \(n_1 = 2\) (as Balmer series transitions end at the second energy level) and \(n_2 = 3\) (as this is the initial energy level for the first Balmer line). Plugging these values into the Rydberg formula, we get: \(\bar\nu = R \times ( \frac{1}{2^2} - \frac{1}{3^2})\)
03

Simplify the expression

Simplify the expression obtained in step 2: \(\bar\nu = R \times ( \frac{1}{4} - \frac{1}{9})\) To find a common denominator for proceeding with the subtraction, we can use 36 as the denominator for both (as it is the least common multiple of 4 and 9) fractions: \(\bar\nu = R \times ( \frac{9}{36} - \frac{4}{36})\) Continue with the subtraction: \(\bar\nu = R \times \frac{9-4}{36}\) \(\bar\nu = R \times \frac{5}{36}\) Thus, the wave number of the first Balmer line in terms of the Rydberg constant R is: \(\boxed{\bar\nu = \frac{5R}{36}}\) This corresponds to option (A).

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