Chapter 6, Problem 34QP

Consider the following energy levels of a hypothetical atom

$\begin{array}{l}{E}_{\text{4}}\text{:-1}{\text{.0×10}}^{\text{-19}}\text{J }{E}_{\text{2}}{\text{: -10×10}}^{\text{-19}}\text{J}\\ E_{\text{3}}\text{:-5}{\text{.0×10}}^{\text{-19}}\text{J }{E}_{\text{1}}{\text{:-15×10}}^{\text{-19}}\text{ J}\end{array}$

(a) What is the wavelength of the photon needed to excite an electron from $E_{\text{1}}\text{to }{E}_{\text{4}}$ ? (b) What is the energy (in joules) a photon must have to excite an electron from $E_{\text{2}}\text{to }{E}_{\text{3}}$ ? (c) When an electron drops from the $E_3$ level to the $E_1$ level, the atom is said to undergo emission. Calculate the wavelength of the photon emitted in this process.

Interpretation Introduction

Interpretation:

The wavelength of photon needed to excite an electron from state $E_1\text{ to }{E}_4$, the energy of photon to excite an electron (in joules) from $E_2\text{ to }{E}_3$, and the wavelength of a photon emitted when an electron drops from $E_3\text{ to }{E}_1$ are to be determined.

Concept introduction:

The relationship between wavelength and frequency of an electromagnetic radiation is as follows:

$\text{c}=\lambda \times \nu$

Here, $\text{c}$ is the speed of light $\left(3.00\times {10}^8\text{ m/s}\right)$, $\lambda$ is the wavelength, and $\nu$ is frequency.

The energy of a photon can be expressed as follows:

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

Here, E is the energy of photon, $\text{h}$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$, $\text{c}$ is the speed of light $\left(3.0\times {10}^8\text{ m/s}\right)$, and $\lambda$ is the wavelength.

The energy difference when an electron drops from $n_f^{}$ to $n_i$ state can be evaluated as follows:

$\begin{array}{c}\Delta E=E_f-E_n\\ =-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\end{array}$

Here, $\Delta E$ is the energy difference, $E_f$ is the energy of the electron in the final energy state, $E_n$ is the energy of the electron in the initial energy state, $n_f^{}$ is the final energy state, and $n_i^{}$ is the initial energy state.

Conversion of meter ($\text{m}$) to nanometer $\text{nm}$ is $\frac{1.0\text{ nm}}{1.0\times {10}^{-9}\text{ m}}$.

Answer to Problem 34QP

Solution:

(a)

$1.4\times {10}^2\text{ nm}$

(b)

$5.0\times {10}^{-19}\text{ J}$

(c)

$2.0\times {10}^2\text{ nm}$

Explanation of Solution

Given information:

$\begin{array}{c}{E}_1=-15\times {10}^{-19}\text{ J}\\ E_4=-1.0\times {10}^{-19}\text{ J}\end{array}$

$\begin{array}{c}{E}_2=-10\times {10}^{-19}\text{ J}\\ E_3=-5.0\times {10}^{-19}\text{ J}\end{array}$

a) The wavelength of the photon needed to excite an electron from $E_1\text{ to }{E}_4$.

The energy difference between states $E_1\text{ to }{E}_4$ can be calculated as follows:

$\begin{array}{c}\Delta E=E_4-E_1\\ =-1.0\times {10}^{-19}\text{ J}-\left(-15\times {10}^{-19}\text{ J}\right)\\ =14\times {10}^{-19}\text{ J}\end{array}$

The wavelength of photon needed to excite an electron from $E_1\text{ to }{E}_4$ can be evaluated as follows:

$\begin{array}{c}\lambda =\frac{\text{hc}}{\Delta E}\\ =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{ m/s}\right)}{14\times {10}^{-19}\text{ J}}\left(\frac{1.0\times {10}^9\text{ nm}}{\text{1 m}}\right)\\ =1.4\times {10}^2\text{ nm}\end{array}$

Hence, the wavelength is $1.4\times {10}^2\text{ nm}$.

b) The energy (in joules) a photon must have to excite an electron from $E_2\text{ to }{E}_3$.

The energy (in joules) required to excite an electron from $E_2\text{ to }{E}_3$ is evaluated as follows:

$\begin{array}{c}\Delta E=E_3-E_2\\ =-5.0\times {10}^{-19}\text{ J}-\left(-10\times {10}^{-19}\text{ J}\right)\\ =5.0\times {10}^{-19}\text{ J}\end{array}$

Therefore, the required energy is $5.0\times {10}^{-19}\text{ J}$.

c) The wavelength of a photon emitted when an electron drops from $E_3\text{ to }{E}_1$.

The wavelength of a photon emitted when an electron drops from $E_3\text{ to }{E}_1$ can be evaluated as follows:

$\begin{array}{c}\Delta E=E_1-E_3\\ \frac{hc}{\lambda }=-15\times {10}^{-19}\text{ J}-\left(-5.0\times {10}^{-19}\text{ J}\right)\\ =-10\times {10}^{-19}\text{ J}\end{array}$

Thus, the energy difference is $-10\times {10}^{-19}\text{ J}$, the negative sign indicates that energy is emitted out.

The wavelength of a photon emitted when an electron drops from state $E_3\text{ to }{E}_1$ can be evaluated by ignoring the negative sign of $\Delta E$:

$\begin{array}{c}\lambda =\frac{\text{hc}}{\Delta E}\\ =\frac{\left(6.63\times {10}^{-34}\cancel{\text{J}}\cdot \cancel{\text{s}}\right)\left(3.0\times {10}^8\cancel{\text{m}}\text{/}\cancel{\text{s}}\right)}{10\times {10}^{-19}\cancel{\text{J}}}\left(\frac{1.0\times {10}^9\text{ nm}}{\text{1 }\cancel{\text{m}}}\right)\\ =2.0\times {10}^2\text{ nm}\end{array}$

Hence, the wavelength is $2.0\times {10}^2\text{ nm}$.