Chapter 30, Problem 6TP

To determine

(a)

The frequency of photon absorbed by the atom.

Answer to Problem 6TP

The frequency of photon absorbed by the atom is $4.59\times {10}^{14}\text{Hz}$.

Explanation of Solution

Given:

The energy of the level 2 is $E_2=-3.4\text{eV}$.

The energy of the level 3 is $E_3=-1.5\text{eV}$.

Formula used:

The change in energy between initial and final orbit is given by

  $\Delta E=E_3-E_2$

The frequency of photon absorbed by the atom is given by

  $\begin{array}{c}\upsilon =\frac{\Delta E}{h}\\ =\frac{E_3-E_2}{h}\end{array}$

Calculation:

The planks constant is $h=6.626\times {10}^{-34}\text{J}\cdot \text{s}$.

The frequency of photon absorbed by the atom is calculated as follows:

  $\begin{array}{c}\upsilon =\frac{E_3-E_2}{h}\\ =\frac{\left(-1.5\text{eV}\right)-\left(-3.4\text{eV}\right)}{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}\\ =\frac{\left(1.9\text{eV}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}\\ =\frac{\left(3.04\times {10}^{-19}\text{J}\right)}{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}\end{array}$

  $\begin{array}{c}\upsilon =\left(4.59\times {10}^{14}{\text{s}}^{\text{-1}}\right)\left(\frac{1\text{Hz}}{1{\text{s}}^{\text{-1}}}\right)\\ =4.59\times {10}^{14}\text{Hz}\end{array}$

Conclusion:

Thus, the frequency of photon absorbed by the atom is $4.59\times {10}^{14}\text{Hz}$.

To determine

(b)

The frequency of photon emitted by the atom.

Answer to Problem 6TP

The frequency of photon emitted by the atom is $2.42\times {10}^{15}\text{Hz}$.

Explanation of Solution

Given:

The energy of the level 1 is $E_1=-13.4\text{eV}$.

Formula used:

The change in energy between initial and final orbit is given by

  $\Delta E=E_2-E_1$

The frequency of photon emitted by the atom is given by

  $\begin{array}{c}\upsilon =\frac{\Delta E}{h}\\ =\frac{E_2-E_1}{h}\end{array}$

Calculation:

The frequency of photon emitted by the atom is calculated as follows:

  $\begin{array}{c}\upsilon =\frac{E_2-E_1}{h}\\ =\frac{\left(-3.4\text{eV}\right)-\left(-13.4\text{eV}\right)}{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}\\ =\frac{\left(10\text{eV}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}\\ =\frac{\left(16\times {10}^{-19}\text{J}\right)}{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}\end{array}$

  $\begin{array}{c}\upsilon =\left(2.42\times {10}^{15}{\text{s}}^{\text{-1}}\right)\left(\frac{1\text{Hz}}{1{\text{s}}^{\text{-1}}}\right)\\ =2.42\times {10}^{15}\text{Hz}\end{array}$

Conclusion:

Thus, the frequency of photon emitted by the atom is $2.42\times {10}^{15}\text{Hz}$.