Question 1.45 - Chemistry: Molecular Nature Of Matter And Change

Question

The distance between two adjacent peaks on a wave is called the wavelength. (a) The wavelength of a beam of ultraviolet light is 247 nanometres (nm). What is its wavelength in meters? (b) The wavelength of a beam of red light is 6760 pm. What is its wavelength in angstroms( $\mathbf{\AA }$ )?

Step-by-Step Solution

Answer

Answer

 

1. The wavelength of the UV beam in meters is $2.47 \times {10}^{-7} \mathrm{m}$.
2. The wavelength of the beam of red light in angstrom is $67.6 \mathrm{\AA }$.

Part (a)

A UV beam has a wavelength of  $\mathrm{\lambda } =247 \mathrm{nm}$. To convert the value to nanometres, use the formula $\mathbf{1}\mathbf{nm}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{m}$ .

$$\begin{gathered} \lambda = 247nm \times \left(\frac{{10}^{-9}m}{1nm}\right) \\ = 247nm \times {10}^{-9}m \\ = 2.47 \times {10}^{-7}m \end{gathered}$$

Hence, 

the wavelength of the UV beam in meters

 is $\mathbf{2}\mathbf{.}\mathbf{47}\mathbf{ }\mathbf{\times }\mathbf{ }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{m}$ .

Part (b)

 The wavelength of a red light beam is $\mathrm{\lambda } = 6760 \mathrm{pm}$ . Convert the value in Pico meters to angstroms using the following formula:

$$\begin{gathered} 1pm ={10}^{-12}m, {10}^{10} \AA =1m \\ \lambda = 6760pm\times \left(\frac{{10}^{-12}m}{1pm}\right)\times \left(\frac{{10}^{10} \AA }{1m}\right) \\ = 6760 \times {10}^{-12}\times {10}^{10} \AA \\ = 67.6 \AA \end{gathered}$$

 

Hence, the wavelength of the beam of red light in angstrom is $\mathbf{67}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{\AA }$ .