Question

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

Answer 1

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,
• $\frac{1}{\lambda }=R{Z}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Where,

• $\lambda$ is the wavelength
• $R$ is the Rydberg constant and has a value $1.097\times {10}^7{m}^{-1}$
• $Z$ is the atomic number
• $n_1$and $n_2$ are energy levels where $n_2>n_1$

Here it is given that the $\lambda$ 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 $n_1$ will be equal to $2$ and $n_2$ 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,

$$\begin{gathered} \frac{1}{\lambda }=R{Z}^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ =R{Z}^2\left[\frac{1}{4}-\frac{1}{9}\right] \\ =\frac{5}{36}R{Z}^2 \end{gathered}$$

• Rearrange this equation to get the value of $R{Z}^2$

$$\begin{gathered} \frac{1}{\lambda }=\frac{5}{36}R{Z}^2 \\ R{Z}^2=\frac{36}{5\lambda } \end{gathered}$$

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

• The first member of the Lyman series means that $n_1=1$and $n_2=2$. Let's assume that the wavelength of the first member of the Lyman series is ${\lambda }_L$.
• Substituting these values in the Rydberg equation, we get,

$$\begin{gathered} \frac{1}{{\lambda }_L}=R{Z}^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\ =R{Z}^2\left[\frac{1}{1}-\frac{1}{4}\right] \\ =\frac{3}{4}R{Z}^2 \\ {\lambda }_L=\frac{4}{3R{Z}^2} \end{gathered}$$

• Now substitute the value of $R{Z}^2$ that we got from step 1 calculation.

$$\begin{gathered} {\lambda }_L=\frac{4}{3R{Z}^2} \\ =\frac{4}{3\times {\displaystyle \frac{36}{5\lambda }}} \\ =\frac{4\times 5\lambda }{3\times 36} \\ =\frac{5\lambda }{27} \end{gathered}$$

Therefore, the wavelength of the first member of the Lyman series is $\frac{\mathbf{5}\mathbf{\lambda }}{\mathbf{27}}$.