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Question

The wavelength of the first member of the Balmer series in the Hydrogen series is λ. Calculate the wavelength of the first member of the Lyman series in the same spectrum.


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Solution

Step 1: Wavelength of the first member of the Balmer series.

  • We can use Rydberg's formula to compute the wavelength. The formula is given below,
  • 1λ=RZ21n12-1n22

Where,

  • λ is the wavelength
  • R is the Rydberg constant and has a value 1.097×107m-1
  • Z is the atomic number
  • n1and n2 are energy levels where n2>n1

Here it is given that the λ is the wavelength for the first member of the Balmer series.

  • From the figure given below, we can understand that the first member of the Balmer series starts from n=2.

  • Thus here n1 will be equal to 2 and n2 will be equal to 3. And the Z is 1 here since the atomic number of hydrogen is 1.
  • Substitute these values in the above equation we get,

1λ=RZ2122-132=RZ214-19=536RZ2

  • Rearrange this equation to get the value of RZ2

1λ=536RZ2RZ2=365λ

Step 2: Wavelength of the first member of the Lyman series.

  • The first member of the Lyman series means that n1=1and n2=2. Let's assume that the wavelength of the first member of the Lyman series is λL.
  • Substituting these values in the Rydberg equation, we get,

1λL=RZ2112-122=RZ211-14=34RZ2λL=43RZ2

  • Now substitute the value of RZ2 that we got from step 1 calculation.

λL=43RZ2=43×365λ=4×5λ3×36=5λ27

Therefore, the wavelength of the first member of the Lyman series is 5λ27.


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