Question

How many wavelengths are emitted by atomic hydrogen in visible range (380 nm − 780 nm)? In the range 50 nm to 100 nm?

Answer 1

Balmer series contains wavelengths ranging from 364 nm (for n₂ = 3) to 655 nm (n₂ = $\infty$).
So, the given range of wavelength (380−780 nm) lies in the Balmer series.
The wavelength in the Balmer series can be found by
$\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$
Here, R = Rydberg's constant = 1.097×10⁷ m^$-$1

The wavelength for the transition from n = 3 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ {\lambda }_1=656.3\mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 4 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ {\lambda }_2=486.1\mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 5 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_3}=R\left(\frac{1}{2^2}-\frac{1}{5^2}\right) \\ {\lambda }_3=434.0\mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 6 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_4}=R\left(\frac{1}{2^2}-\frac{1}{6^2}\right) \\ {\lambda }_4=410.2\mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 7 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_5}=R\left(\frac{1}{2^2}-\frac{1}{7^2}\right) \\ {\lambda }_5=397.0\mathrm{nm} \end{gathered}$$

Thus, the wavelengths emitted by the atomic hydrogen in visible range (380−780 nm) are 5.
Lyman series contains wavelengths ranging from 91 nm (for n₂ = 2) to 121 nm (n₂ =$\infty$).
So, the wavelengths in the given range (50−100 nm) must lie in the Lyman series.
The wavelength in the Lyman series can be found by
$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$

The wavelength for the transition from n = 2 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_1}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ {\lambda }_1=122\mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 3 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_2}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ {\lambda }_2=103\mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 4 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_3}=R\left(\frac{1}{1^2}-\frac{1}{4^2}\right) \\ {\lambda }_3=97.3\mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 5 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_4}=R\left(\frac{1}{1^2}-\frac{1}{5^2}\right) \\ {\lambda }_4=95.0\mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 6 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_5}=R\left(\frac{1}{1^2}-\frac{1}{6^2}\right) \\ {\lambda }_5=93.8\mathrm{nm} \end{gathered}$$

So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.