Chapter 5, Problem 25QRT

(a)

Interpretation Introduction

Interpretation:

Flame tests depend on emission in the visible region of the spectrum to identify elements in sample.  The compounds containing Barium emit at $\text{493}\text{nm}$; strontium compounds emits at $\text{642}\text{nm}$.  The identifying color of the flame test in each case has to be determined.

Explanation of Solution

Given, compounds containing barium emits at the wave length $\text{493}\text{nm}$.

• The color of the wave length $\text{493}\text{nm}$, is blue-green color.

Barium emits blue-green color in flame test.

Compounds containing strontium emits at the wave length $\text{642}\text{nm}$.

• The color of the wave length $\text{642}\text{nm}$, is orange-red color.

Strontium emits orange-red color in flame test.

(b)

Interpretation Introduction

Interpretation:

Flame tests depend on emission in the visible region of the spectrum to identify elements in sample.  The compounds containing Barium emit at $\text{493}\text{nm}$; strontium compounds emits at $\text{642}\text{nm}$.  The energy associated with each of these emissions have to be calculated.

Explanation of Solution

The wave length of barium is given as $\text{493}\text{nm}$.

The energy associated with the emission is calculated as

  $\begin{array}{l}{\text{E}}_{\text{photon}}\text{=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\left(\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{s}\right)\text{×}\left(\text{2}\text{.998}\text{×}{\text{10}}^{\text{8}}\text{m/s}\right)}{\text{493}\text{nm}}\text{×}\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\\ \text{=}\text{4}\text{.03}\text{×}{\text{10}}^{\text{-19}}\text{J}\end{array}$

The wave length of strontium is given as $\text{642}\text{nm}$.

The energy associated with the emission is calculated as

  $\begin{array}{l}{\text{E}}_{\text{photon}}\text{=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\left(\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{s}\right)\text{×}\left(\text{2}\text{.998}\text{×}{\text{10}}^{\text{8}}\text{m/s}\right)}{\text{642}\text{nm}}\text{×}\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\\ \text{=}\text{3}\text{.09}\text{×}{\text{10}}^{\text{-19}}\text{J}\text{.}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The reason has to be explained for the emission of barium at lower wave length than the strontium does.

Explanation of Solution

The relationship between energy and wave length is

  $\text{E}\text{=}\frac{\text{hc}}{\text{λ}}$

It states that energy and wave length are inversely proportional.

The energy emitted by barium at the wave length of $\text{493}\text{nm}$ is $\text{4}\text{.03}\text{×}{\text{10}}^{\text{-19}}\text{J}$.

The energy emitted by strontium at the wave length of $\text{642}\text{nm}$ is $\text{3}\text{.09}\text{×}{\text{10}}^{\text{-19}}\text{J}$.

Hence, higher the energy, lower the wave length.  Barium consists of higher energy, so it has lower wave length than strontium.