Question

The wavelength of first member of balmer series is 6563 Angstroms. Calculate the wavelength of second member of lyman series.

Answer 1

Lyman series start from n1 = 1 and n2=2,3, 4 , 5,.....

so, second member of Lyman series
n1 = 1 and n2=3

now,
1/lemda =R.Z^2[1/n1^2 -1/n2^2]
=RZ^2[1/1^2 -1/3^2] ------(1)

now,
given,
first member of balmer series

1/6563 =R.Z^2[1/2^2 - 1/3^2] --------(2)

divide equation (1) and (2)

6563/lemda =(1-1/9)/(1/4-1/9)=(8/9)/(5/36)
=32/5

lemda =6563 × 5/32 =1025.48 A°

Answer 2

The wavelength of second member of lyman series is $1215 \times 10^{-10}=1215 \AA$

Given data:

First member of the Balmer series has wavelength of 6563.

$\lambda^{\prime \prime}=6563 \AA$

Solution:

Wavelength of spectral lines are derived from the formula for the hydrogen spectrum, which is given below:

$\frac{1}{\lambda}=R\left(\frac{1}{n_{f^{2}}}-\frac{1}{n_{i}^{2}}\right)$

Where,

R as the Rydberg constant

We know that, the Balmer series member $n_f$ and $n_i$ values are 2 and 3 respectively.

Therefore on solving the formula with the values, we get,

$-\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$

$-\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{9}\right)$

$-\frac{1}{\lambda}=R\left(\frac{9-4}{36}\right)$

$\Rightarrow \frac{1}{\lambda}=\frac{5 R}{36}$

Now, for the Lyman series, $n_f$ and $n_i$ values are 1 and 2 respectively.

On substituting on the formula given above, we get,

$\Rightarrow \frac{1}{\lambda^{n}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$

$\Rightarrow \frac{1}{\lambda^{\prime \prime}}=\frac{3 R}{4}$

Thereby, on dividing the equation we have for the Balmer and Lyman series, we get,

$\frac{\lambda}{\lambda^{\prime \prime}}=\left(\frac{5 R}{36}\right) \times\left(\frac{4}{3 R}\right)$

$\Rightarrow \frac{\lambda}{\lambda^{\prime \prime}}=\frac{5}{27}$

$\Rightarrow \lambda=\frac{5}{27} \lambda^{\prime \prime}$

On substituting, the wavelength given in questions, we get,

$\lambda=\frac{5 \times 6563 \times 10^{-10}}{27}$

$\lambda=1215 \times 10^{-10}=1215 \AA$