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The ratio of shortest wavelength lines i...

The ratio of shortest wavelength lines in Lyman , Balmer and Paschen series is 1:4:x. Calculate the value of x ?

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To solve the problem of finding the value of \( x \) in the ratio of the shortest wavelength lines in the Lyman, Balmer, and Paschen series, we will follow these steps: ### Step 1: Understand the Series and Their Shortest Wavelengths - The Lyman series corresponds to transitions to the first energy level (n=1). - The Balmer series corresponds to transitions to the second energy level (n=2). - The Paschen series corresponds to transitions to the third energy level (n=3). ### Step 2: Use the Rydberg Formula The Rydberg formula for the wavelength of emitted light is given by: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level. ### Step 3: Calculate the Shortest Wavelength for Each Series **For the Lyman Series (n=1):** - The shortest wavelength occurs when \( n_2 \) approaches infinity. - Thus, using the Rydberg formula: \[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - 0 \right) = R \] So, \( \lambda_L = \frac{1}{R} \). **For the Balmer Series (n=2):** - The shortest wavelength occurs when \( n_2 \) approaches infinity. - Thus: \[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4} \] So, \( \lambda_B = \frac{4}{R} \). **For the Paschen Series (n=3):** - The shortest wavelength occurs when \( n_2 \) approaches infinity. - Thus: \[ \frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} \] So, \( \lambda_P = \frac{9}{R} \). ### Step 4: Establish the Ratios From the calculations, we have: - \( \lambda_L = \frac{1}{R} \) - \( \lambda_B = \frac{4}{R} \) - \( \lambda_P = \frac{9}{R} \) The ratios of the wavelengths are: \[ \lambda_L : \lambda_B : \lambda_P = \frac{1}{R} : \frac{4}{R} : \frac{9}{R} = 1 : 4 : 9 \] ### Step 5: Relate to Given Ratio The problem states that the ratio is \( 1 : 4 : x \). From our calculations, we found that the ratio is \( 1 : 4 : 9 \). Thus, we can conclude that: \[ x = 9 \] ### Final Answer The value of \( x \) is \( 9 \). ---

To solve the problem of finding the value of \( x \) in the ratio of the shortest wavelength lines in the Lyman, Balmer, and Paschen series, we will follow these steps: ### Step 1: Understand the Series and Their Shortest Wavelengths - The Lyman series corresponds to transitions to the first energy level (n=1). - The Balmer series corresponds to transitions to the second energy level (n=2). - The Paschen series corresponds to transitions to the third energy level (n=3). ### Step 2: Use the Rydberg Formula ...
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Calculate the ratio of shortest wavelength possible in Lyman and Balmer series.

Show that the shortest wavelength lines in Lyman,Balmer and Paschen series have their wavelength in the ratio 1:4:9.

Knowledge Check

  • The ratio of the longest to the shortest wavelength lines in the Balmer series is

    A
    1.1
    B
    8.8
    C
    1.8
    D
    8.1
  • What is the shortest wavelength line in the Paschen series of Li^(2+) ion ?

    A
    `(R)/(9)`
    B
    `(9)/(R)`
    C
    `(1)/(R)`
    D
    `(9R)/(4)`
  • The shortest wavelength of Lyman series is 912 Å . That of paschen series is

    A
    8460 Å
    B
    8208 Å
    C
    8415 Å
    D
    8430 Å
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