Chapter 1, Problem 1B.8E

(a)

Interpretation Introduction

Interpretation:

The energy of an excited copper atom when it generates an x - ray photon of frequency $1.2\times {10}^{17}Hz$ has to be calculated.

Concept Introduction:

Energy of a photon can be expressed mathematically as given below.

  $E=h\nu =\frac{hc}{\lambda }$

Where, h is the Planck’s constant, $\nu$ is the frequency of radiation, $c$ is the speed of light and $\lambda$ is the wavelength of the radiation.

Answer to Problem 1B.8E

The energy of an excited copper atom when it generates an x - ray photon of frequency $1.2\times {10}^{17}Hz$ is $7.95\times 10^{-17}J.$

Explanation of Solution

Given that, the frequency is $1.2\times {10}^{17}Hz.$

The energy of an excited copper atom when it generates an x - ray photon of frequency $1.2\times {10}^{17}Hz$ can be calculated as given below.

  $\begin{array}{l}E=h\nu \\ \\ E=\left(6.626\times {10}^{-34}J.s\right)\times \left(1.2\times {10}^{17}{s}^{-1}\right)\\ \\ E=7.95\times 10^{-17}J.\end{array}$

Therefore, the energy of an excited copper atom when it generates an x - ray photon of frequency $1.2\times {10}^{17}Hz$ is $7.95\times 10^{-17}J.$

(b)

Interpretation Introduction

Interpretation:

The energy of $2.00mole$ of excited copper atoms emitting light of frequency $1.2\times {10}^{17}Hz$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 1B.8E

The energy of $2.00mole$ of excited copper atoms emitting light of frequency $1.2\times {10}^{17}Hz$ is $9.56\times 10^{7}J.mo{l}^{-1}.$

Explanation of Solution

The energy of one copper atom is $7.95\times 10^{-17}J.$

The energy of $2.00mole$ of excited copper atoms emitting light of frequency $1.2\times {10}^{17}Hz$ can be calculated as shown below.

  $\begin{array}{l}E\left(permoleofphotons\right)=N_{A}E\left(1photon\right)\\ \\ E\left(permoleofphotons\right)=\left(6.022\times {10}^{23}mo{l}^{-1}\right)\times \left(\text{7}{\text{.95×10}}^{\text{-17}}\text{J}\right)\\ \\ E\left(permoleofphotons\right)=4.7{\text{8×10}}^{\text{7}}\text{J}{\text{.mol}}^{\text{-1}}\\ \\ E\left(2.00moleofphotons\right)=2\times 4.7{\text{8×10}}^{\text{7}}\text{J}{\text{.mol}}^{\text{-1}}=9.56\times 10^{7}J.mo{l}^{-1}\end{array}$

Therefore, the energy of $2.00mole$ of excited copper atoms emitting light of frequency $1.2\times {10}^{17}Hz$ is $9.56\times 10^{7}J.mo{l}^{-1}.$

(c)

Interpretation Introduction

Interpretation:

The energy of $2.00g$ of copper atoms emitting light of frequency $1.2\times {10}^{17}Hz$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 1B.8E

The energy of $2.00g$ of copper atoms emitting light of frequency $1.2\times {10}^{17}Hz$ is $1.50\times 10^{6}J.$

Explanation of Solution

Calculation of number of copper atoms present in $2.00g$ of copper atoms:

  $\begin{array}{l}1moleofCu=63.546g\\ \\ 1moleofCucontains6.022\times {10}^{23}atoms.\\ \\ So,2.00g\times \frac{1mole}{63.546g}\times \frac{6.022\times {10}^{23}atoms}{1mole}=0.189\times 10^{23}atoms\end{array}$

The energy of one copper atom is $7.95\times 10^{-17}J.$  Hence, the energy of $0.189\times 10^{23}atoms$ of copper can be calculated as given below.

  $E=\left(0.189\times {10}^{23}\right)\times \left(\text{7}{\text{.95×10}}^{\text{-17}}\text{J}\right)=1.50\times 10^{6}J$

Therefore, the energy of $2.00g$ of copper atoms emitting light of frequency $1.2\times {10}^{17}Hz$ is $1.50\times 10^{6}J.$