Chapter 5, Problem 5.1P

To determine

To Find: The wavelength of the light emitted by the hydrogen and its nature.

Answer to Problem 5.1P

  $\lambda =1.87\text{μm}$, Infrared.

Explanation of Solution

Given:

Given $n=4$ and $n^1=3$

Formula used:

Rydberg formula:

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

Where R is the Rydberg constant $R=0.0110{\text{nm}}^{-1}$

Calculation:

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

  $\frac{1}{\lambda }=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$

  $\frac{1}{\lambda }=\left(0.0110{\text{nm}}^{-1}\right)\left(\frac{1}{9}-\frac{1}{16}\right)$

  $\frac{1}{\lambda }=\frac{7}{16\times 9}\times 0.0110{\text{nm}}^{-1}$

  $\frac{1}{\lambda }=5.3472\times {10}^{-4}{\text{nm}}^{-1}$

  $\Rightarrow \lambda =1870.12\text{nm}$

  $\lambda =1.87\mu \text{m}$

Conclusion:

Hence, the wavelength is $\lambda =1.87\text{μm}$ and it lies in the infrared region.