Q14.141CP - Chemistry: Molecular Nature Of Matter And Change

Question

The electronic transition in Na from 3p¹ to 3s¹ gives rise to a bright yellow-orange emission at 589.2 nm. What is the energy of this transition?

Step-by-Step Solution

Answer

The energy of this transition is $\text{3.371}\times {\text{10}}^{\text{ - 19 }} \text{J}$.

Step 1: Convert wavelengths in nanometres (nm) to metres (m).

The following equation is used to determine the energy of the transition:

$E=\frac{hc}{\lambda }\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left(1\right)$

E is the energy of transition, h is the Plank's constant, c is the light velocity and  $\lambda$ is the emission wavelength.

Plank's constant has a value of $6.26\times {10}^{-34} \mathrm{J}.\mathrm{s}$ .

The value of light velocity is $2.99792\times {10}^8 \mathrm{m}/\mathrm{s}$.

The emission wavelength is 589.2 nm.

Convert wavelengths in nanometres (nm) to metres (m).

$\begin{array}{rcl}\lambda & =& 589.2 \text{nm}\times \frac{1 \text{m}}{{10}^9 \text{nm}}\\ & =& 589.2\times {10}^{-9} \text{m}\\ & & \end{array}$

Step 2: Evaluation of energy of transition

Substitute all known values in equation (1).

$\begin{array}{rcl}E& =& \frac{\left(6.26\times {10}^{-34} \text{J.s}\right)\times \left(2.99792\times {10}^8 \text{m/s}\right)}{\left(589.2\times {10}^{-9} \text{m}\right)}\\ & =& 3.371\times {10}^{-19} \text{J}\\ & & \end{array}$

Therefore, the energy of transition is $\begin{array}{rcl}& & 3.371\times {10}^{-19} \text{J}\end{array}$ .