Question

By calculating its wavelength, show that the first line in the Lyman series is UV radiation.

Short answer

It is proved that the first line in the Lyman series is UV radiation.

Step-by-step solution

Determine the formulas:

Consider the formula for the charge to mass ratio in the electron and the proton as follows:

$$\begin{array}{c}\frac{q_e}{m_e}=-1.76\times {10}^{11}\frac{\mathrm{C}}{\mathrm{k}\mathrm{g}}\\ \frac{q_p}{m_p}=9.57\times {10}^7\frac{\mathrm{C}}{\mathrm{k}\mathrm{g}}\end{array}$$

Here,$$m_e=9.11\times {10}^{-31}\mathrm{k}\mathrm{g}$$is the mass of the electron and$$m_p=1.67\times {10}^{-27}\mathrm{k}\mathrm{g}$$is the mass of the proton.

Consider the formula for the Bohr's theory of hydrogen atom.

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

Here, wavelength of the emitted electromagnetic radiation is λ , and the Rydberg constant is $$R=1.097\times {10}^7{\mathrm{m}}^{-1}$$.

Determine the wavelength of the Layman series:

Consider the given values:

$$\begin{array}{c}{n}_i=2\\ n_f=1\\ R=1.097\times {10}^7{\mathrm{m}}^{-1}\end{array}$$

Substitute the values and solve as:

$$\begin{array}{c}\frac{1}{\lambda }=1.097\times {10}^7{\mathrm{m}}^{-1}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\\ \frac{1}{\lambda }=8.2275\times {10}^6{\mathrm{m}}^{-1}\\ \lambda =1.215\times {10}^{-7}\mathrm{m}\end{array}$$

Wavelength is 122 nm.

The wavelength of the layman series is$$1.215\times {10}^{-7}\mathrm{m}$$.