Nature of radiation Questions in English

(A) Given,wavelength $\lambda = 456 \, nm = 456 \times 10^{-9} \, m$.
Speed of light $c = 3 \times 10^8 \, m/s$.
The relationship between frequency $\nu$,speed of light $c$,and wavelength $\lambda$ is given by the formula: $c = \nu \lambda$.
Rearranging for frequency: $\nu = \frac{c}{\lambda}$.
Substituting the values: $\nu = \frac{3 \times 10^8 \, m/s}{456 \times 10^{-9} \, m}$.
$\nu = 0.0065789 \times 10^{17} \, s^{-1} = 6.579 \times 10^{14} \, Hz$.
Rounding to two decimal places,we get $\nu \approx 6.58 \times 10^{14} \, Hz$.

(A) The relationship between speed of light $(c)$,wavelength $(\lambda)$,and frequency $(\nu)$ is given by: $c = \lambda \nu$
Rearranging the formula to solve for wavelength: $\lambda = \frac{c}{\nu}$
Given: $c = 3 \times 10^8 \ m \ s^{-1}$ and $\nu = 5.09 \times 10^{14} \ s^{-1}$
$\lambda = \frac{3 \times 10^8 \ m \ s^{-1}}{5.09 \times 10^{14} \ s^{-1}}$
$\lambda = 5.894 \times 10^{-7} \ m$
Converting to nanometers: $\lambda = 589.4 \times 10^{-9} \ m = 589.4 \ nm$

(A) The relationship between wavelength $(\lambda)$,frequency $(\nu)$,and speed of light $(c)$ is given by $\lambda = \frac{c}{\nu}$.
Given:
Speed of light $(c) = 3 \times 10^8 \ m/s$
Frequency $(\nu) = 1368 \ kHz = 1368 \times 10^3 \ Hz$
Calculation:
$\lambda = \frac{3 \times 10^8 \ m/s}{1368 \times 10^3 \ s^{-1}}$
$\lambda = \frac{300000000}{1368000} \ m$
$\lambda \approx 219.298 \ m$
Rounding to one decimal place,we get $\lambda = 219.3 \ m$.

(N/A) The wavelength $\lambda$ is calculated using the formula $\lambda = \frac{c}{\nu}$,where $c = 3.0 \times 10^8 \ m \ s^{-1}$ and $\nu = 4.620 \times 10^{14} \ Hz$.
$\lambda = \frac{3.0 \times 10^8 \ m \ s^{-1}}{4.620 \times 10^{14} \ Hz} = 0.6494 \times 10^{-6} \ m$.
Converting to nanometers: $\lambda = 0.6494 \times 10^{-6} \ m \times 10^9 \ nm/m = 649.4 \ nm$.
The visible spectrum range is approximately $400 \ nm$ to $750 \ nm$. Since $649.4 \ nm$ falls within this range,it belongs to the visible region.

(B) Given: Frequency $(\nu) = 4 \times 10^{11} \ kHz = 4 \times 10^{14} \ Hz$ (or $s^{-1}$).
Speed of light $(c) = 3 \times 10^8 \ m \ s^{-1}$.
We know the relation: $\nu = \frac{c}{\lambda}$.
Therefore,wavelength $(\lambda) = \frac{c}{\nu}$.
$\lambda = \frac{3 \times 10^8 \ m \ s^{-1}}{4 \times 10^{14} \ s^{-1}}$.
$\lambda = 0.75 \times 10^{-6} \ m = 7.5 \times 10^{-7} \ m$.

(A) The relationship between frequency $(f)$,wavelength $(\lambda)$,and speed of light $(c)$ is given by the equation:
$f \times \lambda = c$
Given:
$\lambda = 1.5 \times 10^{-10} \ m$
$c = 3 \times 10^8 \ m \ s^{-1}$
Rearranging the formula to solve for frequency $(f)$:
$f = \frac{c}{\lambda}$
Substituting the values:
$f = \frac{3 \times 10^8 \ m \ s^{-1}}{1.5 \times 10^{-10} \ m} = 2 \times 10^{18} \ s^{-1} = 2 \times 10^{18} \ Hz$

(N/A) Electromagnetic radiation exhibits dual behavior,meaning it possesses both wave-like and particle-like properties.
$1$. Wave-like nature: This is supported by phenomena such as interference and diffraction.
$2$. Particle-like nature: This is supported by phenomena such as black body radiation and the photoelectric effect.
Thus,light behaves as a wave in some contexts and as a stream of particles (photons) in others.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Since $h$ and $c$ are constants,the energy is inversely proportional to the wavelength: $E \propto \frac{1}{\lambda}$.
Therefore,the ratio of energies $E_1 : E_2$ for wavelengths $\lambda_1 = 6000 \ \mathring{A}$ and $\lambda_2 = 4000 \ \mathring{A}$ is:
$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = \frac{4000}{6000} = \frac{4}{6} = \frac{2}{3}$.
Thus,the ratio is $2:3$.

The energy of a photon is given by $E = \frac{hc}{\lambda}$.
For $\lambda_1 = 7000 \ \mathring{A} = 7000 \times 10^{-10} \ m$:
$E_1 = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{7000 \times 10^{-10}} = 2.837 \times 10^{-19} \ J$.
For $\lambda_2 = 400 \ nm = 400 \times 10^{-9} \ m$:
$E_2 = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.965 \times 10^{-19} \ J$.

(A) Given: Periodic time $T = 4 \times 10^{-10} \ s$.
Frequency $\nu = \frac{1}{T} = \frac{1}{4 \times 10^{-10}} = 0.25 \times 10^{10} = 2.5 \times 10^{9} \ Hz$.
Wavelength $\lambda = c \times T = (3 \times 10^{8} \ m/s) \times (4 \times 10^{-10} \ s) = 12 \times 10^{-2} \ m = 0.12 \ m$.
Wave number $\bar{\nu} = \frac{1}{\lambda} = \frac{1}{0.12} = 8.33 \ m^{-1}$.

(N/A) Absorption spectra: When electromagnetic radiation interacts with an object,it can be absorbed by atoms,molecules,or ions,causing them to transition from a lower energy state $E_{1}$ to a higher energy state $E_{2}$. The energy absorbed is given by $\Delta E = E_{2} - E_{1} = h\nu$. This process results in an absorption spectrum,which appears as dark lines in a continuous spectrum.
Emission spectra: Atoms or molecules in an excited state $E_{2}$ are unstable and tend to return to a lower energy state $E_{1}$ by releasing energy in the form of electromagnetic radiation. The energy emitted is given by $\Delta E = E_{2} - E_{1} = h\nu$. This process results in an emission spectrum,which appears as bright lines on a dark background.
Spectroscopy: The study of emission or absorption spectra is referred to as spectroscopy. In the gaseous phase,these spectra are not continuous but consist of discrete lines.

(N/A) Line spectra (atomic spectra): The emission spectra of atoms in the gas phase do not show a continuous spread of wavelength from red to violet. Instead,they emit light only at specific wavelengths with dark spaces between them. Such spectra are called line spectra or atomic spectra because the emitted radiation is identified by the appearance of bright lines in the spectra.
An emission or absorption spectrum is a photographic recording of the separated wavelengths,which is referred to as a line spectrum.
Uses: Line emission spectra are of great interest in the study of electronic structure. The characteristic lines in atomic spectra can be used in chemical analysis to identify unknown atoms. By matching the lines of the emission spectrum of the atoms of a known element with the lines from an unknown sample,one can identify the elements present. Elements like $Rb$,$Cs$,$Tl$,$In$,$Ga$,and $Sc$ were discovered when their minerals were analyzed by spectroscopic methods.

(N/A) The mass of a photon is given by the relation $m = \frac{h}{\lambda c}$.
Given: $h = 6.626 \times 10^{-34} \ J \cdot s$,$\lambda = 5890 \ \mathring{A} = 5890 \times 10^{-10} \ m$,$c = 3 \times 10^8 \ m/s$.
$m = \frac{6.626 \times 10^{-34}}{5890 \times 10^{-10} \times 3 \times 10^8} = 3.75 \times 10^{-36} \ kg$.
The mass of an electron is $m_e = 9.11 \times 10^{-31} \ kg$.
Ratio $= \frac{m}{m_e} = \frac{3.75 \times 10^{-36}}{9.11 \times 10^{-31}} \approx 4.11 \times 10^{-6}$ times.

(B < A < C = D) The energy of a radiation is given by $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.
First,convert all wavelengths to meters:
$\lambda(A) = 300 \ nm = 300 \times 10^{-9} \ m = 3 \times 10^{-7} \ m$
$\lambda(B) = 300 \ \mu m = 300 \times 10^{-6} \ m = 3 \times 10^{-4} \ m$
$\lambda(C) = 3 \ nm = 3 \times 10^{-9} \ m$
$\lambda(D) = 30 \ \mathring{A} = 30 \times 10^{-10} \ m = 3 \times 10^{-9} \ m$
Comparing the wavelengths: $\lambda(B) > \lambda(A) > \lambda(C) = \lambda(D)$.
Since energy is inversely proportional to wavelength,the increasing order of energy is $B < A < C = D$.

(N/A) The wavelength $(\lambda)$ is the distance between two successive peaks or two successive troughs of a wave.
In the given figure, the distance between a peak and the next equilibrium point is $2.16 \text{ pm}$.
Since a full wavelength corresponds to $4$ such segments (peak to equilibrium, equilibrium to trough, trough to equilibrium, equilibrium to peak), the wavelength is:
$\lambda = 4 \times 2.16 \text{ pm} = 8.64 \text{ pm}$

(N/A) The wavelength $\lambda$ is calculated using the formula $\lambda = \frac{c}{\nu}$.
Given: $c = 3.0 \times 10^{8} \ m \ s^{-1}$ and $\nu = 4.620 \times 10^{14} \ Hz$.
$\lambda = \frac{3.0 \times 10^{8} \ m \ s^{-1}}{4.620 \times 10^{14} \ s^{-1}} = 0.6494 \times 10^{-6} \ m$.
Converting to nanometers: $0.6494 \times 10^{-6} \ m \times 10^{9} \ nm/m = 649.4 \ nm$.
Since the wavelength $649.4 \ nm$ falls within the range of $400 \ nm$ to $750 \ nm$,it belongs to the visible light region of the electromagnetic spectrum.

(N/A) Photoelectric effect: When radiation with a certain minimum frequency $v_{0}$ strikes the surface of a metal,electrons are ejected from the metal surface. This phenomenon is called the photoelectric effect. The emitted electrons are called photoelectrons.
Equipment for studying the photoelectric effect: Light of a particular frequency strikes a clean metal surface inside a vacuum chamber. Electrons are ejected from the metal and are counted by a detector that measures their kinetic energy.
The results observed in this experiment were:
$(i)$. The electrons are ejected from the metal surface as soon as the beam of light strikes the surface,i.e.,there is no time lag between the striking of the light beam and the ejection of electrons from the metal surface.
$(ii)$. The number of electrons ejected is proportional to the intensity or brightness of light.
$(iii)$. For each metal,there is a characteristic minimum frequency,$v_{0}$ (also known as threshold frequency),below which the photoelectric effect is not observed. At a frequency $v > v_{0}$,the ejected electrons come out with certain kinetic energy. The kinetic energies of these electrons increase with the increase of the frequency of the light used.
The above observations cannot be explained by the electromagnetic wave theory. According to this theory,since radiations were continuous,it should be possible to accumulate energy on the surface of the metal,irrespective of its frequency,and thus,radiations of all frequencies should be able to eject electrons.
Similarly,according to this theory,the energy of the ejected electrons should depend upon the intensity of the incident radiation.

(B) $X$-rays are electromagnetic radiations that do not deflect in electric or magnetic fields,indicating they are neutral.
They possess a very short wavelength (approximately $0.1 \ nm$).
Due to their high penetrating power,they are widely used in medical imaging to study the internal structure of objects and the human body.

(B) Scattering is the phenomenon of the deflection of a wave due to an obstacle. It can be explained by the wave nature of electromagnetic radiation. When electromagnetic radiation interacts with particles of matter,the waves are redirected in various directions,which is known as scattering.

(A) Interference is a phenomenon that occurs when two or more electromagnetic waves overlap in space.
When the crests and troughs of the waves align,they undergo constructive interference,resulting in an increased amplitude.
When the crest of one wave aligns with the trough of another,they undergo destructive interference,resulting in a decreased amplitude.
This behavior is a fundamental property of waves,demonstrating the wave nature of electromagnetic radiation.

(A) The wave nature of electromagnetic radiation explains phenomena such as diffraction and interference.
The particle nature of electromagnetic radiation explains phenomena such as the photoelectric effect and black body radiation.

(D) In the visible spectrum,the violet color has the maximum energy because it has the shortest wavelength.
For violet light:
Wavelength $(\lambda)$: Approximately $380-450 \, nm$ (or $3800-4500 \, \mathring{A}$).
Frequency $(\nu)$: Approximately $6.7 \times 10^{14} - 7.9 \times 10^{14} \, Hz$.
Energy $(E = h\nu)$: It has the highest energy among visible colors,approximately $2.75 - 3.26 \, eV$ per photon.

(A) In the visible spectrum,the energy of light is inversely proportional to its wavelength $(E = \frac{hc}{\lambda})$.
Red color has the longest wavelength (approximately $700 \ nm$),which corresponds to the minimum energy.
Using $E = \frac{hc}{\lambda}$ where $h = 6.626 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9}} \approx 2.84 \times 10^{-19} \ J$.
The frequency $\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{700 \times 10^{-9}} \approx 4.28 \times 10^{14} \ Hz$.

(N/A) According to this theory,when a charged particle is accelerated,it emits electromagnetic radiation.

(N/A) The different regions of the electromagnetic spectrum are: Radio waves,Microwave region,Infrared region,Visible region,Ultraviolet region,$X$-rays,and $\gamma$-rays.

(N/A) Electromagnetic radiation is characterized by the following properties:
$(i)$ Wavelength $(\lambda)$
$(ii)$ Frequency $(\nu)$
$(iii)$ Wavenumber $(\bar{\nu})$

(A) In a vacuum,all types of electromagnetic radiation travel at a constant speed of $3.0 \times 10^{8} \text{ m s}^{-1}$.
This speed is known as the speed of light and is denoted by the symbol $c$.

(N/A) Wave number $= \bar{v} = \frac{1}{\lambda} \text{ m}^{-1}$ (It is the number of wavelengths per unit length.)
Time period $= T = \frac{1}{v} \text{ s}$ (The time taken by a wave to complete one oscillation or cycle is called the time period $(T)$.)

(N/A) As iron is heated,its color changes with temperature: $Dull \ Red$ $\rightarrow Bright \ Red$ $\rightarrow White$ $\rightarrow Blue$. As the temperature increases,the wavelength of the emitted radiation decreases,which means the frequency $(v)$ of the emitted radiation increases.

(B) black body is an idealized physical body that absorbs all incident electromagnetic radiation,regardless of frequency or angle of incidence.
It does not reflect or transmit any radiation.
Because it absorbs all radiation,it is also a perfect emitter of radiation when heated,following Planck's law.

(A) The intensity of radiation is defined as the amount of energy flowing per unit area per unit time. In terms of the particle nature of light,it corresponds to the number of photons passing through a unit area per unit time. Higher intensity means a greater number of photons are present in the beam.

(N/A) Visible Spectrum: White light contains waves of all wavelengths in the visible region. When a beam of white light is passed through a prism,it splits into a series of colored bands,which is called a 'spectrum'.
Continuous Spectrum: The visible spectrum ranges from violet $(7.50 \times 10^{14} \ Hz)$ to red $(4.00 \times 10^{14} \ Hz)$. This is a continuous spectrum. $A$ 'continuous spectrum' is one in which the adjacent colors (wavelengths) blend into each other without any gaps.
Example: In the visible spectrum,violet blends into blue,blue into green,and so on,with all colors merging into one another.

(A) The examples of a continuous spectrum are:
$(i)$ The rainbow.
$(ii)$ The spectrum obtained by passing a beam of white light through a prism.

(A) When a beam of light passes through a medium,it undergoes refraction and deviates from its path.
Different wavelengths of light travel at different speeds in a medium,causing them to bend by different angles.
Shorter wavelengths (higher frequency) bend more than longer wavelengths.
This separation of light into its constituent colors is known as dispersion,which results in the formation of a spectrum.

(A) Violet light undergoes greater refraction than red light. This is because violet light has a shorter wavelength and higher frequency compared to red light.

Color	Wavelength $(\lambda)$	Frequency $(v)$	Deviation
Violet	$400 \, nm$	$7.50 \times 10^{14} \, Hz$	Greater
Red	$750 \, nm$	$4.00 \times 10^{14} \, Hz$	Lesser

(B) An emission spectrum is produced when radiation emitted from a source (like an excited atom or molecule) is passed through a prism or diffraction grating.
It consists of specific wavelengths or frequencies of light that are characteristic of the substance being studied.
This spectrum provides information about the energy levels of atoms and molecules.

(N/A) To observe an electron,it is illuminated with light or electromagnetic radiation having a wavelength smaller than the dimensions of the electron. This allows the interaction of photons with the electron to be detected.

(N/A) $(i)$ $10^{-8}$
$(ii)$ $10^6$
$(iii)$ Hertz $(Hz)$
$(iv)$ Wave number $(\overline{\nu})$

(N/A) The wave number $(\bar{\nu})$ is defined as the number of wavelengths per unit length. Its $SI$ unit is $m^{-1}$,and it is commonly expressed in $cm^{-1}$.

(D) The frequency ranges for electromagnetic radiation are as follows:
$(1)$ $X$-rays have high frequency,typically around $10^{18} \ Hz$ $(D)$.
$(2)$ Ultraviolet $(UV)$ rays have frequencies around $10^{16} \ Hz$ $(C)$.
$(3)$ Long radio waves have very low frequencies,typically $10^0 - 10^4 \ Hz$ $(A)$.
$(4)$ Microwaves have frequencies around $10^{10} \ Hz$ $(B)$.
Therefore,the correct matching is: $(1-D, 2-C, 3-A, 4-B)$.

(A) The spectral lines of the Balmer series for the $H$-atom correspond to transitions from higher energy levels to the $n = 2$ energy level. These transitions emit radiation that falls within the visible region of the electromagnetic spectrum.

(D) The energy $E$ required to ionize one mole of atoms is given by the formula:
$E = \frac{hcN_{A}}{\lambda \times 1000} \text{ (in } kJ \ mol^{-1} \text{)}$
Given:
$h = 6.63 \times 10^{-34} \ J \cdot s$
$c = 3.00 \times 10^{8} \ m \cdot s^{-1}$
$N_{A} = 6.02 \times 10^{23} \ mol^{-1}$
$\lambda = 663 \ nm = 663 \times 10^{-9} \ m$
Substituting the values:
$E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8} \times 6.02 \times 10^{23}}{663 \times 10^{-9} \times 1000}$
$E = \frac{6.63 \times 3.00 \times 6.02 \times 10^{-3}}{663 \times 10^{-6}}$
$E = \frac{119.799 \times 10^{-3}}{663 \times 10^{-6}} = \frac{119.799}{0.663} \approx 180.69 \ kJ \ mol^{-1}$
Rounding off to the nearest integer,we get $181 \ kJ \ mol^{-1}$.

(A) The relationship between frequency $(\nu)$,wavelength $(\lambda)$,and speed of light $(c)$ is given by the formula: $\nu = \frac{c}{\lambda}$.
Rearranging the formula to solve for wavelength: $\lambda = \frac{c}{\nu}$.
Given: $c = 3.0 \times 10^{8} \, m \, s^{-1}$ and $\nu = 1,368 \, kHz = 1,368 \times 10^{3} \, Hz$.
Substituting the values: $\lambda = \frac{3.0 \times 10^{8} \, m \, s^{-1}}{1,368 \times 10^{3} \, s^{-1}}$.
$\lambda = \frac{300,000,000}{1,368,000} \, m \approx 219.29 \, m$.
Rounding to one decimal place,we get $\lambda \approx 219.3 \, m$.

(C) The energy of one mole of photons is given by the formula: $E = \frac{hcN_A}{\lambda}$
Substituting the given values:
$E = \frac{6.63 \times 10^{-34} \ J \ s \times 3 \times 10^8 \ m \ s^{-1} \times 6.02 \times 10^{23} \ mol^{-1}}{300 \times 10^{-9} \ m}$
$E = \frac{11.976 \times 10^{-2} \ J \ mol^{-1}}{3 \times 10^{-7}}$
$E = 3.9913 \times 10^5 \ J \ mol^{-1}$
$E \approx 399 \ kJ \ mol^{-1}$

(C) According to Henry Moseley's law for characteristic $X$-rays,the frequency $v$ is related to the atomic number $Z$ by the equation: $\sqrt{v} = a(Z - b)$,where $a$ and $b$ are constants.
This can be written as $v^{1/2} = a(Z - b)$.
Comparing this with the given plot of $v^{n}$ versus $Z$,we get $n = 1/2 = 0.5$.

(B) The energy of one photon is given by $E = h\nu$.
The energy of one mole of photons is given by $E = N_A \times h \times \nu$.
Substituting the given values:
$E = (6.022 \times 10^{23} \ mol^{-1}) \times (6.626 \times 10^{-34} \ Js) \times (2 \times 10^{12} \ s^{-1})$.
$E = 6.022 \times 6.626 \times 2 \times 10^{23-34+12} \ J \ mol^{-1}$.
$E = 79.795 \times 10^{1} \ J \ mol^{-1} = 797.95 \ J \ mol^{-1}$.
Rounding to the nearest integer,we get $798 \ J \ mol^{-1}$.

(D) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.
For one mole of photons,the energy is $E_{mol} = \frac{hcN_A}{\lambda}$.
$E_{mol} = \frac{6.626 \times 10^{-34} \, J \, s \times 3 \times 10^8 \, m \, s^{-1} \times 6.022 \times 10^{23} \, mol^{-1}}{300 \times 10^{-9} \, m} = 3.99 \times 10^5 \, J \, mol^{-1}$.
According to the photoelectric effect equation,$E_{photon} = \Phi + KE$,where $\Phi$ is the work function (minimum energy to remove an electron).
$\Phi = E_{mol} - KE = (3.99 \times 10^5 - 1.68 \times 10^5) \, J \, mol^{-1} = 2.31 \times 10^5 \, J \, mol^{-1}$.

(B) Statement $A$ is correct: Line emission spectra provide information about the electronic structure of atoms.
Statement $B$ is incorrect: Atoms in the gas phase emit line spectra (discrete wavelengths),not a continuous spectrum.
Statement $C$ is correct: An absorption spectrum consists of dark lines in a continuous spectrum,which corresponds to the bright lines in an emission spectrum,acting like a photographic negative.
Statement $D$ is correct: Helium was indeed discovered in the sun's spectrum during a solar eclipse.
Therefore,only $1$ statement (Statement $B$) is incorrect.

(B) black body is an ideal body that can emit and absorb all frequencies of electromagnetic radiation. Thus,statement $(A)$ is correct.
The frequency distribution of the emitted radiation from a black body depends solely on its temperature. Thus,statement $(B)$ is correct.
For a black body at a given temperature,the intensity of emitted radiation varies with frequency,showing a characteristic curve that passes through a maximum value. Thus,statement $(C)$ is correct.
According to Wien's displacement law,as the temperature of a black body increases,the peak of the intensity distribution curve shifts towards higher frequencies (or shorter wavelengths). Thus,statement $(D)$ is correct.
Since all statements $(A)$,$(B)$,$(C)$,and $(D)$ are correct,the number of incorrect statements is $0$.

(D) From the given figure, the distance shown corresponds to half of the wavelength (half-cycle), i.e., $\frac{\lambda}{2} = 1.5 \ pm$.
Therefore, the wavelength $\lambda = 2 \times 1.5 \ pm = 3 \ pm = 3 \times 10^{-12} \ m$.
The relationship between frequency $(\nu)$, wavelength $(\lambda)$, and speed of light $(c)$ is $\nu = \frac{c}{\lambda}$.
Given $c = 3 \times 10^8 \ m/s$, we have $\nu = \frac{3 \times 10^8 \ m/s}{3 \times 10^{-12} \ m} = 1 \times 10^{20} \ Hz$.
Expressing this in the form $x \times 10^{19} \ Hz$, we get $10 \times 10^{19} \ Hz$.
Thus, $x = 10$.