Chapter 7.2, Problem 7.3BFP

(a)

Interpretation Introduction

Interpretation:

The energy difference for the transition of an electron from $n=6$ to a lower energy level for 1 mole of $\text{H}$ atoms is to be determined.

Concept introduction:

Electromagnetic waves are radiations that are formed by oscillating electric and magnetic fields. The electric and magnetic field components of an electromagnetic wave are perpendicular to each other.

The energy of quanta is related to the difference in an atom’s energy states as follows:

$\Delta E=h\nu$        (1)

Here,

$\Delta E$ is the energy difference between two energy levels in an atom.

$h$ is the Plank’s constant.

$\nu$ is the frequency

The equation that relates to the frequency and wavelength of electromagnetic radiation is as follows:

$\nu =\frac{c}{\lambda }$

Here,

$c$ is the speed of light.

$\lambda$ is the wavelength.

Equation (1) can be modified as follows:

$\Delta E=h\left(\frac{c}{\lambda }\right)$        (2)

The conversion factor to convert wavelength from $\text{nm}$ to $\text{m}$ is,

$1\text{nm}=1\times {10}^{-9}\text{ m}$

The conversion factor to convert $\text{kJ}$ to $\text{J}$ is,

$1\text{ kJ}=1000\text{ J}$

(b)

Interpretation Introduction

Interpretation:

The energy level to which the electron moves after it emits a photon of wavelength $410\text{ nm}$ is to be determined.

Concept introduction:

The equation to find the difference in the energy between the two levels in hydrogen-like atoms is,

$\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_{\text{final}}^2}-\frac{1}{n_{\text{initial}}^2}\right)$        (3)

Here,

$\Delta E$ is the difference in the energy between two levels.

$n_{\text{final}}$ is the lower energy level.

$n_{\text{initial}}$ is the higher energy level.