Chapter 7, Problem 125A

Interpretation Introduction

(a)

Interpretation:

The energy of the photon emitted by the mercury atom needs to be determined.

Concept introduction:

The energy of a photon can be determined by calculating the difference of the energy value between the original energy and the energy of the photon after dropping. Energy is expressed in units of Joules.

Answer to Problem 125A

The energy of the photon emitted by the mercury atom $=3.44\times {10}^{-19}\text{J}$

Explanation of Solution

Given that mercury drops from $1.413\times {10}^{-18}$ joules to $1.069\times {10}^{-18}$ joules.

Therefore, the energy of the photon emitted by the mercury atom

$\begin{array}{l}=1.413\times {10}^{-18}-1.069\times {10}^{-18}\\ =3.44\times {10}^{-19}\text{J}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The frequency of the photon emitted by the mercury atom needs to be determined.

Concept introduction:

The frequency of a photon can be calculated as,

$\nu =\frac{E}{h}$

Where, $\nu$ = frequency of photon, $E$ = energy of photon and $h$ = Planck’s constant.

Answer to Problem 125A

The frequency of the photon emitted by the mercury atom is $=5.19\times {10}^{14}{\text{s}}^{\text{-1}}$

Explanation of Solution

Given that mercury drops from $1.413\times {10}^{-18}$ joules to $1.069\times {10}^{-18}$ joules.

The energy of the photon emitted by the mercury atom $=3.44\times {10}^{-19}\text{J}$

The frequency of the photon emitted by the mercury atom can be calculated by using the formula,

$\begin{array}{l}\nu =\frac{E}{h}\\ =\frac{3.44\times {10}^{-19}\text{J}}{6.626\times {10}^{-34}\text{J}\text{.s}}\\ =5.19\times {10}^{14}{\text{s}}^{\text{-1}}\end{array}$

Here, $\nu$ = frequency of photon, $E$ = energy of photon and $h$ = Planck’s constant.

(c)

Interpretation Introduction

Interpretation:

The wavelength of the photon emitted by the mercury atom needs to be determined.

Concept introduction:

The wavelength of a photon can be calculated using the formula,

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

Where, $\lambda$ = wavelength, $c$ = velocity of photon and $v$ = frequency of photon

Answer to Problem 125A

The wavelength of the photon emitted by the mercury atom $=5.78\times {10}^{-7}\text{m}$

Explanation of Solution

The wavelength of the photon emitted by the mercury atom can be calculated by using the formula,

$\begin{array}{l}\lambda =\frac{c}{v}\\ =\frac{3.00\times {10}^8\text{m/s}}{5.19\times {10}^{14}{\text{s}}^{\text{-1}}}\\ =5.78\times {10}^{-7}\text{m}\end{array}$

Here, $\lambda$ = wavelength, $c$ = velocity of photon and $v$ = frequency of photon