Chapter 9, Problem 71QRT

Interpretation Introduction

Interpretation:

The maximum wavelength of the radiation that would be diffracted by $NaCl$ crystal has to be determined.  The frequency of radiation and the energy associated with one photon and one mole of photons of the radiation have to be calculated.  This radiation belongs to which region of the spectrum has to be determined.

Concept Introduction:

The maximum wavelength of the radiation that would be diffracted by a crystal is equal to its unit cell edge length.

Electromagnetic radiation consists of oscillating, perpendicular electric and magnetic fields that travel through space at the same rate.  The frequency of electromagnetic radiation is related to its wavelength by equation as given below.

  $\nu \lambda =c$

Where, $\nu =$ Frequency of electromagnetic radiation, $\lambda =$ Wavelength of electromagnetic radiation and $c=$ Velocity of light.

The energy of radiation is proportional to its frequency of radiation according to the equation as given below.

  $E=h\nu$

Where, $h=$ Planck’s constant $=6.626\times {10}^{-34}Js$

Frequency and energy range of electromagnetic radiation:

	Frequency Range $\left(Hz\right)$	Energy Range $\left(J\right)$
Gamma rays	${10}^{20}-{10}^{24}$	${10}^{-14}-{10}^{-10}$
$X-$ rays	${10}^{17}-{10}^{20}$	${10}^{-17}-{10}^{-14}$
UV	${10}^{15}-{10}^{17}$	${10}^{-20}-{10}^{-18}$
IR	${10}^{13}-{10}^{14}$	${10}^{-22}-{10}^{-20}$
Microwaves	$3\times {10}^{11}-{10}^{13}$	${10}^{-25}-{10}^{-22}$
Radiowaves	$<3\times {10}^{11}$	${10}^{-30}-{10}^{-25}$

Answer to Problem 71QRT

The frequency of radiation of one photon and one mole photons are found out to be $5.30\times 10^{17}{s}^{-1}$ and $31.91\times 10^{40}{s}^{-1}$ respectively.  The energy associated with one photon and one mole of photons are found to be $3.51\times 10^{-16}J$ and $2.11\times 10^{8}J$ respectively.

Explanation of Solution

The maximum wavelength of the radiation that would be diffracted by a crystal is equal to its unit cell edge length.  So ${\lambda }_{\max }$ for $NaCl$ is $566pm$ which is the unit cell edge length.

(a) Then, the frequency of radiation can be calculated as given below.

  $\nu =\frac{c}{\lambda }=\frac{2.998\times {10}^{8}m/s}{566pm\times \frac{{10}^{-12}m}{1pm}}=5.30\times {10}^{17}{s}^{-1}.$

Therefore, the frequency of radiation of one photon is found out to be $5.30\times 10^{17}{s}^{-1}.$

The energy associated with one photon can also be calculated as follows,

  $E=h\nu =\left(6.626\times {10}^{-34}Js\right)\times \left(5.30\times {10}^{17}{s}^{-1}\right)=3.51\times 10^{-16}J.$

(b) The frequency of radiation of one mole photons can be calculated as follows,

  $1molphoton\times \frac{\text{5}{\text{.30×10}}^{\text{17}}{\text{s}}^{\text{-1}}}{1photon}\times \frac{6.022\times {10}^{23}photons}{1molphoton}=31.91\times 10^{40}{s}^{-1}.$

The energy associated with one mole photons can also be calculated as follows,

  $E=h\nu =\left(6.626\times {10}^{-34}Js\right)\times \left(31.91\times {10}^{40}{s}^{-1}\right)=2.11\times 10^{8}J.$

The energy and frequency range are coming under $X-$ray region of electromagnetic spectrum.