Particle Nature of Light : Photon Questions in English

(B) The concept that photons carry energy is fundamentally linked to the interaction of radiation with matter. While $Bohr's$ theory describes energy levels in atoms,the experimental evidence that photons possess momentum and energy,and can transfer this energy to electrons,was definitively established by the $Compton$ effect. In the $Compton$ effect,an $X$-ray photon collides with an electron,transferring a portion of its energy and momentum to the electron,which confirms the particle-like nature of light and the energy-carrying capacity of photons.

(B) The positron is the antiparticle of the electron.
By definition,an antiparticle has the same mass as its corresponding particle.
Therefore,the mass of a positron is exactly equal to the mass of an electron,which is approximately $9.11 \times 10^{-31} \; kg$ or $0.000548 \; a.m.u.$
Thus,the correct option is $(b)$.

(A) The rest mass energy of a particle is given by Einstein's mass-energy equivalence formula: $E = m_e c^2$.
Here,the rest mass of an electron is $m_e = 9.109 \times 10^{-31} \, kg$ and the speed of light is $c \approx 3 \times 10^8 \, m/s$.
Calculating the energy in Joules:
$E = (9.109 \times 10^{-31} \, kg) \times (3 \times 10^8 \, m/s)^2 \approx 8.198 \times 10^{-14} \, J$.
To convert this energy into electron-volts $(eV)$,we divide by the charge of an electron $(1.602 \times 10^{-19} \, C)$:
$E = \frac{8.198 \times 10^{-14} \, J}{1.602 \times 10^{-19} \, J/eV} \approx 511,735 \, eV \approx 0.511 \, MeV = 511 \, keV$.
Rounding to the standard value,the rest energy of an electron is approximately $510 \, keV$.

(C) The antiparticle of an electron,denoted as $_{-1}e^{0}$,is a positron,denoted as $_{+1}e^{0}$.
Both particles have the same mass,but the positron carries a positive elementary charge,whereas the electron carries a negative elementary charge.

(B) The process of pair production involves a gamma ray photon interacting with a nucleus to produce an electron and a positron.
The energy of the gamma ray photon $(E_{\gamma})$ is used to provide the rest mass energy of the electron $(m_e c^2)$,the rest mass energy of the positron $(m_p c^2)$,and the total kinetic energy $(K.E.)$ of the pair.
Given:
Rest mass energy of electron $(m_e c^2)$ = $0.5\, MeV$.
Rest mass energy of positron $(m_p c^2)$ = $0.5\, MeV$ (since it is the antiparticle of the electron).
Total kinetic energy $(K.E.)$ = $0.78\, MeV$.
Therefore,the energy of the gamma ray photon is:
$E_{\gamma} = m_e c^2 + m_p c^2 + K.E.$
$E_{\gamma} = 0.5\, MeV + 0.5\, MeV + 0.78\, MeV = 1.78\, MeV$.

(C) The electron and the positron are anti-particles of each other.
By definition,a particle and its corresponding anti-particle possess the same mass and spin,but opposite charges.
Therefore,the positron has the same mass as the electron,which is approximately $9.11 \times 10^{-31} \ kg$.

(C) The rest energy of the electron is given by $E_0 = m_0 c^2 = 0.511 \, MeV$.
When the electron is accelerated to a velocity $v = 0.5 \, c$,its total energy $E$ is given by the relativistic formula:
$E = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}$
Substituting $v = 0.5 \, c$:
$E = \frac{0.511}{\sqrt{1 - (0.5)^2}} = \frac{0.511}{\sqrt{1 - 0.25}} = \frac{0.511}{\sqrt{0.75}}$
$E = \frac{0.511}{0.866} \approx 0.590 \, MeV$
The change in energy $\Delta E$ is the difference between the total energy and the rest energy:
$\Delta E = E - E_0 = 0.590 \, MeV - 0.511 \, MeV = 0.079 \, MeV$.

(A) Given,rest mass energy $E_0 = m_0 c^2 = 0.54 \, MeV$.
Velocity of the electron $v = 0.8 \, c$.
The relativistic mass $m$ is given by $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$.
Substituting $v = 0.8 \, c$:
$m = \frac{m_0}{\sqrt{1 - (0.8)^2}} = \frac{m_0}{\sqrt{1 - 0.64}} = \frac{m_0}{\sqrt{0.36}} = \frac{m_0}{0.6}$.
The total energy $E = mc^2 = \frac{m_0 c^2}{0.6}$.
Substituting $m_0 c^2 = 0.54 \, MeV$:
$E = \frac{0.54}{0.6} = 0.9 \, MeV$.
The kinetic energy $K.E.$ is given by $K.E. = E - E_0$.
$K.E. = 0.9 \, MeV - 0.54 \, MeV = 0.36 \, MeV$.

(C) According to the corpuscular theory of light proposed by Isaac Newton,light consists of tiny,massless particles called corpuscles.
Newton postulated that the different colours of light are caused by the different sizes of these corpuscles.
Therefore,the correct option is $C$.

(C) The quantum nature of light,proposed by Max Planck,was introduced to explain the radiation spectrum of a black body. Classical physics (wave theory) failed to explain the energy distribution of black body radiation at shorter wavelengths (the ultraviolet catastrophe). Planck hypothesized that energy is emitted or absorbed in discrete packets called quanta or photons,which successfully explained the observed spectrum.

(B) When an $X$-ray photon collides with a proton,it undergoes Compton scattering.
In this process,the photon transfers some of its energy to the proton (recoil).
Since the energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency,a loss in energy results in a decrease in frequency.
Therefore,the frequency of the scattered photon decreases.

(D) photon is a quantum of electromagnetic radiation. It is an elementary particle that carries energy and momentum,but it has zero rest mass and zero electric charge. Since it has no charge,it does not possess an intrinsic magnetic moment. Therefore,a photon is associated with neither electric charge nor magnetic moment.

(D) The rest mass energy of an electron $(e^-)$ is $0.51 \, MeV$.
Since a positron $(e^+)$ has the same rest mass as an electron,its rest mass energy is also $0.51 \, MeV$.
For pair production,the photon must provide energy equal to the sum of the rest mass energies of the electron and the positron.
Minimum energy $E = m_e c^2 + m_p c^2 = 0.51 \, MeV + 0.51 \, MeV = 1.02 \, MeV$.

(D) The momentum $p$ of a photon is given by the de Broglie relation: $p = \frac{h}{\lambda}$.
Given:
Planck's constant $h = 6.626 \times 10^{-34} \ J \cdot s$.
Wavelength $\lambda = 0.010 \ \mathring A = 0.010 \times 10^{-10} \ m = 10^{-12} \ m$.
Substituting the values:
$p = \frac{6.626 \times 10^{-34}}{10^{-12}} \ kg \cdot m/s$.
$p = 6.626 \times 10^{-22} \ kg \cdot m/s$.

(A) The power $P$ of a source is given by $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.
Thus,the number of photons emitted per second is $n = \frac{P \lambda}{hc}$.
Given $P_{UV} = P_{IR} = 130 \ W$,$\lambda_{UV} = 400 \ nm$,and $\lambda_{IR} = 700 \ nm$.
The ratio of the number of photons emitted per second is:
$\frac{n_{UV}}{n_{IR}} = \frac{P_{UV} \lambda_{UV} / hc}{P_{IR} \lambda_{IR} / hc} = \frac{\lambda_{UV}}{\lambda_{IR}}$.
Substituting the values:
$\frac{n_{UV}}{n_{IR}} = \frac{400 \ nm}{700 \ nm} = \frac{4}{7} \approx 0.57$.

(C) The power $P$ of a source is given by $P = n \cdot E_{photon}$,where $n$ is the number of photons emitted per second and $E_{photon} = \frac{hc}{\lambda}$.
Therefore,the number of photons emitted per second is $n = \frac{P \cdot \lambda}{hc}$.
Since the power $P$ and the constants $h$ and $c$ are the same for all three sources,the number of photons $n$ is directly proportional to the wavelength $\lambda$ $(n \propto \lambda)$.
Comparing the wavelengths: $\lambda_{UV} \approx 400 \ nm$,$\lambda_{Visible} \approx 550 \ nm$,and $\lambda_{IR} \approx 700 \ nm$.
Since $\lambda_{IR} > \lambda_{Visible} > \lambda_{UV}$,the number of photons emitted per second is maximum for the infrared source.

(C) The power $P$ of the bulb is given by the total energy emitted per second,which is $P = \frac{nE}{\Delta t}$,where $n$ is the number of photons per second and $E = \frac{hc}{\lambda}$ is the energy of one photon.
Given:
Power $P = 100 \ W = 100 \ J/s$
Wavelength $\lambda = 540 \ nm = 540 \times 10^{-9} \ m$
Planck's constant $h = 6 \times 10^{-34} \ J \cdot s$
Speed of light $c = 3 \times 10^8 \ m/s$
Substituting these values into the formula $P = \frac{n \cdot hc}{\lambda \cdot t}$:
$100 = \frac{n \times (6 \times 10^{-34}) \times (3 \times 10^8)}{540 \times 10^{-9}}$
Solving for $n$:
$n = \frac{100 \times 540 \times 10^{-9}}{6 \times 10^{-34} \times 3 \times 10^8}$
$n = \frac{54000 \times 10^{-9}}{18 \times 10^{-26}}$
$n = 3000 \times 10^{17} = 3 \times 10^{20}$
Thus,the number of photons emitted per second is $3 \times 10^{20}$.

(A) The energy of a single photon of green light is given by $E = \frac{hc}{\lambda} = \frac{12400 \ eV \cdot \mathring{A}}{5000 \ \mathring{A}} = 2.48 \ eV$.
Converting this to Joules: $E \approx 2.48 \times 1.6 \times 10^{-19} \ J \approx 4 \times 10^{-19} \ J$.
The minimum intensity the eye can detect is $I_{Eye} = (5 \times 10^4 \ photons/m^2 \cdot s) \times (4 \times 10^{-19} \ J/photon) = 2 \times 10^{-14} \ W/m^2$.
The ratio of sensitivity is given by the ratio of the intensities: $\frac{S_{Eye}}{S_{Ear}} = \frac{I_{Ear}}{I_{Eye}} = \frac{10^{-13} \ W/m^2}{2 \times 10^{-14} \ W/m^2} = 5$.
Thus,the eye is $5$ times more sensitive than the ear.

(C) The momentum $p$ of a photon is given by the formula $p = \frac{h}{\lambda}$.
Since the speed of light $c = \nu \lambda$,we have $\lambda = \frac{c}{\nu}$.
Substituting this into the momentum formula,we get $p = \frac{h\nu}{c}$.
Given values are frequency $\nu = 10^9 \ Hz$,Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$,and speed of light $c = 3 \times 10^8 \ m/s$.
$p = \frac{(6.6 \times 10^{-34} \ J \cdot s) \times (10^9 \ Hz)}{3 \times 10^8 \ m/s}$.
$p = \frac{6.6}{3} \times 10^{-34 + 9 - 8} \ kg \ m/s$.
$p = 2.2 \times 10^{-33} \ kg \ m/s$.

(A) Given: Frequency $f = 880 \ kHz = 880 \times 10^3 \ Hz$, Power $P = 10 \ kW = 10^4 \ J/s$.
The energy of a single photon is given by $E_{photon} = hf$, where $h = 6.626 \times 10^{-34} \ J \cdot s$.
Let $n$ be the number of photons emitted per second. The total power $P$ is equal to the energy emitted per second, so $P = n \times E_{photon} = n \times hf$.
Rearranging for $n$: $n = \frac{P}{hf} = \frac{10^4}{(6.626 \times 10^{-34}) \times (880 \times 10^3)}$.
$n = \frac{10^4}{5.83 \times 10^{-28}} \approx 1.715 \times 10^{31} \approx 1.72 \times 10^{31}$ photons/second.

(C) The power of the bulb is $P = 60 \ W$,which means $60 \ J$ of energy is emitted per second.
Let $n$ be the number of photons emitted per second.
The energy of one photon is given by $E_{photon} = \frac{hc}{\lambda}$.
The total energy emitted per second is $E = n \times E_{photon} = n \frac{hc}{\lambda}$.
Rearranging for $n$,we get $n = \frac{E \lambda}{hc}$.
Substituting the given values: $E = 60 \ J$,$\lambda = 660 \times 10^{-9} \ m$,$h = 6.6 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$.
$n = \frac{60 \times 660 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{39600 \times 10^{-9}}{19.8 \times 10^{-26}} = 2000 \times 10^{17} = 2 \times 10^{20}$ photons/second.

(B) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Using the shortcut formula $E \approx \frac{12400 \ \text{eV} \cdot \mathring{A}}{\lambda (\text{in } \mathring{A})}$,we substitute the given wavelength:
$E = \frac{12400 \ \text{eV} \cdot \mathring{A}}{4000 \ \mathring{A}}$
$E = 3.1 \ \text{eV}$.

(C) The energy of a single photon is given by $E = h\nu$,where $h = 6.63 \times 10^{-34} \ J \cdot s$ and $\nu = 6 \times 10^{14} \ Hz$.
$E = (6.63 \times 10^{-34}) \times (6 \times 10^{14}) \approx 3.98 \times 10^{-19} \ J$.
The power $P$ entering the eye is given by $P = I \times A$,where $I = 10^{-10} \ W/m^2$ and $A = 0.4 \ cm^2 = 0.4 \times 10^{-4} \ m^2 = 4 \times 10^{-5} \ m^2$.
$P = 10^{-10} \times 4 \times 10^{-5} = 4 \times 10^{-15} \ W$.
The number of photons per second $n$ is given by $n = P / E$.
$n = (4 \times 10^{-15}) / (3.98 \times 10^{-19}) \approx 1.005 \times 10^4 \approx 10^4$ photons per second.

(A) Given: Power $P = 100 \ W$,Efficiency $\eta = 5\% = 0.05$,Wavelength $\lambda = 5.6 \times 10^{-5} \ cm = 5.6 \times 10^{-7} \ m$.
The power emitted as visible light is $P' = P \times \eta = 100 \times 0.05 = 5 \ W$.
The energy of one photon is $E = \frac{hc}{\lambda}$.
The number of photons (quanta) emitted per second $n$ is given by $n = \frac{P'}{E} = \frac{P' \lambda}{hc}$.
Substituting the values: $n = \frac{5 \times 5.6 \times 10^{-7}}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{28 \times 10^{-7}}{19.89 \times 10^{-26}} \approx 1.407 \times 10^{19} \text{ quanta/sec}$.

(A) The power $P$ of the transmitter is given by $P = nE$,where $n$ is the number of photons emitted per second and $E$ is the energy of a single photon.
Energy of a single photon is $E = \frac{hc}{\lambda}$.
Therefore,$P = n \frac{hc}{\lambda}$,which implies $n = \frac{P\lambda}{hc}$.
Given: $P = 10 \ kW = 10^4 \ W$,$\lambda = 500 \ m$,$h = 6.626 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$.
Substituting the values:
$n = \frac{10^4 \times 500}{6.626 \times 10^{-34} \times 3 \times 10^8}$
$n = \frac{5 \times 10^6}{19.878 \times 10^{-26}}$
$n \approx 0.2515 \times 10^{32} \ s^{-1} = 2.515 \times 10^{31} \ s^{-1}$.
Rounding to the nearest option,we get $2.5 \times 10^{31} \ s^{-1}$.

(C) Given wavelength $\lambda = 6.20 \times 10^{-5} \ cm = 6.20 \times 10^{-7} \ m = 620 \ nm$.
The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Using the value $hc \approx 1240 \ eV \cdot nm$,we get:
$E = \frac{1240 \ eV \cdot nm}{620 \ nm} = 2.0 \ eV$.
Therefore,the energy of the photon is $2 \ eV$.

(B) The energy of a photon is given by $E = hv$.
According to Einstein's mass-energy equivalence and the relation between energy and momentum for a photon,$E = pc$,where $p$ is the momentum and $c$ is the speed of light.
Therefore,the momentum $p$ is given by $p = E/c$.
Substituting the value of $E$,we get $p = hv/c$.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Since $h$ (Planck's constant) and $c$ (speed of light) are constants,the energy $E$ is inversely proportional to the wavelength $\lambda$,i.e.,$E \propto \frac{1}{\lambda}$.
Therefore,the ratio of the energies is given by $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Substituting the given values: $\frac{E_1}{E_2} = \frac{300 \ nm}{150 \ nm} = 2$.
Thus,the ratio is $2:1$ or simply $2$.

(C) According to the theory of special relativity,the relativistic mass $m$ of an object moving with velocity $v$ is given by the formula:
$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$
Given that the speed $v = 0.8 c$,we substitute this value into the equation:
$m = \frac{m_0}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}}$
$m = \frac{m_0}{\sqrt{1 - 0.64}}$
$m = \frac{m_0}{\sqrt{0.36}}$
$m = \frac{m_0}{0.6}$
$m = \frac{m_0}{6/10} = \frac{10}{6} m_0 = \frac{5}{3} m_0$
Therefore,the mass of the electron at this speed is $\frac{5}{3} m_0$.

(B) The momentum of a photon is given by $p = E/c$.
Since the energy of a photon is $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Substituting $E$ in the momentum equation: $p = (h\nu)/c$.
Rearranging for frequency $\nu$: $\nu = (pc)/h$.

(C) The power of the bulb is $P = 100 \ W$,which means it emits $100 \ J$ of energy per second.
The energy of a single photon is given by the formula $E = \frac{hc}{\lambda}$.
Given values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $\lambda = 410 \times 10^{-9} \ m$.
Substituting these values,the energy of one photon is $E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{410 \times 10^{-9}} \approx 4.85 \times 10^{-19} \ J$.
The number of photons emitted per second $(n)$ is calculated as $n = \frac{P}{E}$.
$n = \frac{100}{4.85 \times 10^{-19}} \approx 2.06 \times 10^{20}$ photons per second.

(C) Given energy $E = 2 \ cal/cm^2 \cdot min$. Converting to $SI$ units: $E = 2 \times 4.2 \ J/cm^2 \cdot min = 8.4 \ J/cm^2 \cdot min$.
The energy of a single photon is given by $\epsilon = \frac{hc}{\lambda}$.
The total energy $E$ incident on the surface is $E = n \epsilon = n \frac{hc}{\lambda}$,where $n$ is the number of photons.
Rearranging for $n$: $n = \frac{E \lambda}{hc}$.
Substituting the values: $n = \frac{8.4 \times 5500 \times 10^{-10}}{6.6 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{46200 \times 10^{-10}}{19.8 \times 10^{-26}} = \frac{4.62 \times 10^{-6}}{1.98 \times 10^{-25}} \approx 2.33 \times 10^{19}$.
Thus,the number of photons is $2.3 \times 10^{19} \ photons/cm^2 \cdot min$.

(D) The wavelength of the photon is given as $\lambda = 0.01 \ \mathring A = 0.01 \times 10^{-10} \ m = 10^{-12} \ m$.
According to the de Broglie relation,the momentum $p$ of a photon is given by $p = \frac{h}{\lambda}$.
Substituting the value of $\lambda$:
$p = \frac{h}{10^{-12}} = h \times 10^{12} \ kg \ m/s$.

(A) The energy of a photon is given by the formula $E = hf = \frac{hc}{\lambda}$.
Given:
Planck's constant $h = 6.63 \times 10^{-34} \, J \cdot s$
Speed of light $c = 3 \times 10^8 \, m/s$
Wavelength $\lambda = 450 \, nm = 450 \times 10^{-9} \, m$
Substituting the values:
$E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{450 \times 10^{-9}}$
$E = \frac{19.89 \times 10^{-26}}{450 \times 10^{-9}}$
$E = 0.0442 \times 10^{-17} \, J = 4.42 \times 10^{-19} \, J$
Rounding to two significant figures,$E \approx 4.4 \times 10^{-19} \, J$.

(B) The energy of a photon is given by the formula $E = hf$,where $E$ is the energy,$h$ is Planck's constant,and $f$ is the frequency.
First,convert the energy from $eV$ to Joules: $E = 100 \ eV = 100 \times 1.6 \times 10^{-19} \ J = 1.6 \times 10^{-17} \ J$.
Now,rearrange the formula to solve for frequency: $f = \frac{E}{h}$.
Substitute the values: $f = \frac{1.6 \times 10^{-17} \ J}{6.62 \times 10^{-34} \ J \cdot s}$.
$f \approx 0.2417 \times 10^{17} \ Hz$.
$f = 2.417 \times 10^{16} \ Hz$.

$(A)$ The power $P$ emitted by the bulb is given by the total energy of $n$ photons emitted per second: $P = n \cdot E = n \cdot \frac{hc}{\lambda}$.
Rearranging for $n$, we get $n = \frac{P \lambda}{hc}$.
Given values: $P = 60 \, mW = 60 \times 10^{-3} \, W$, $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \, m$, $h = 6.62 \times 10^{-34} \, J \cdot s$, and $c = 3 \times 10^8 \, m/s$.
Substituting these values:
$n = \frac{(60 \times 10^{-3}) \times (6000 \times 10^{-10})}{(6.62 \times 10^{-34}) \times (3 \times 10^8)}$
$n = \frac{360000 \times 10^{-13}}{19.86 \times 10^{-26}}$
$n \approx 1.812 \times 10^{17} \approx 1.8 \times 10^{17} \, \text{photons/second}$.

(D) The Compton effect is the scattering of a photon by a charged particle,usually an electron.
It results in a decrease in energy (increase in wavelength) of the photon,which is called the Compton shift.
This phenomenon cannot be explained by the classical wave theory of light.
It can only be explained by treating light as a stream of particles (photons) that collide with electrons.
In this collision,both energy and momentum are conserved.
Therefore,the Compton effect provides direct evidence that photons possess momentum,confirming the particle nature of electromagnetic radiation.

(C) The energy of a photon is given by $E = hf = \frac{hc}{\lambda}$.
Therefore, $E \propto \frac{1}{\lambda}$.
Given $\lambda_1 = 6000 \ \mathring{A}$, $E_1 = 3.32 \times 10^{-19} \ J$.
Given $\lambda_2 = 4000 \ \mathring{A}$.
Using the ratio: $\frac{E_2}{E_1} = \frac{\lambda_1}{\lambda_2} = \frac{6000}{4000} = 1.5$.
Thus, $E_2 = 1.5 \times E_1 = 1.5 \times 3.32 \times 10^{-19} \ J = 4.98 \times 10^{-19} \ J$.
To convert energy from Joules to $eV$, divide by $1.6 \times 10^{-19} \ J/eV$:
$E_2 (in \ eV) = \frac{4.98 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.1125 \ eV \approx 3.1 \ eV$.

(A) Given: Power $P = 100\, W$,Wavelength $\lambda = 694\, nm = 694 \times 10^{-9}\, m$.
Energy of one photon $E = \frac{hc}{\lambda}$.
The number of photons emitted per second $n$ is given by $n = \frac{P}{E} = \frac{P\lambda}{hc}$.
Substituting the values: $h = 6.63 \times 10^{-34}\, J\cdot s$,$c = 3 \times 10^8\, m/s$.
$n = \frac{100 \times 694 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{69400 \times 10^{-9}}{19.89 \times 10^{-26}} \approx 3489.19 \times 10^{17} \approx 3.49 \times 10^{20}$ photons/second.

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$ (in Joules).
To convert this to $eV$,we divide by the elementary charge $e$: $E_{(eV)} = \frac{hc}{e\lambda}$.
Substituting the values $h = 6.626 \times 10^{-34} \, J\cdot s$,$c = 3 \times 10^8 \, m/s$,and $e = 1.6 \times 10^{-19} \, C$,and expressing $\lambda$ in $\mathring{A}$ $(1 \, \mathring{A} = 10^{-10} \, m)$:
$E_{(eV)} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times \lambda_{(\mathring{A})} \times 10^{-10}} \approx \frac{12400}{\lambda_{(\mathring{A})}} \, eV$.
Since $1 \, KeV = 1000 \, eV$,we have $E_{(KeV)} = \frac{12400}{1000 \times \lambda_{(\mathring{A})}} = \frac{12.4}{\lambda_{(\mathring{A})}}$.
Thus,the energy in $KeV$ is given by $E = \frac{12.4}{\lambda}$.

(C) According to the modern quantum theory,light exhibits a dual nature.
$(1)$ Wave nature: Light behaves as an electromagnetic wave in phenomena like interference,diffraction,and polarization.
$(2)$ Particle nature: Light behaves as a stream of particles called photons in phenomena like the photoelectric effect and Compton scattering.

(A) The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
We know that the speed of light $c$ (or velocity of a photon $v$) is related to frequency and wavelength by $v = \nu\lambda$,so $\nu = v/\lambda$.
Substituting this into the energy equation: $E = h(v/\lambda) = (h/\lambda)v$.
We are given the momentum of the photon as $p = h/\lambda$.
Substituting $p$ into the energy equation: $E = pv$.
Therefore,the velocity of the photon is $v = E/p$.

(A) Given: Frequency $v = 880 \times 10^3 \ Hz$,Power $P = 10 \times 10^3 \ W$.
Let $n$ be the number of photons emitted per second.
The power $P$ is given by $P = n \times E$,where $E = hv$ is the energy of a single photon.
$E = (6.63 \times 10^{-34} \ J \cdot s) \times (880 \times 10^3 \ Hz) = 5.8344 \times 10^{-28} \ J$.
Now,$n = \frac{P}{E} = \frac{10 \times 10^3}{5.8344 \times 10^{-28}}$.
$n \approx 1.71 \times 10^{31} \ \text{photons/second}$.

(D) The energy of a photon is given by $E = hv$,where $h$ is Planck's constant and $v$ is the frequency.
According to Einstein's mass-energy equivalence,$E = mc^2$.
Equating the two,we get $mc^2 = hv$,which implies $mc = hv/c$.
Since momentum $p = mc$,we have $p = hv/c$.

(A) The momentum $p$ of a photon is given by the relation $p = \frac{h}{\lambda}$.
Since the speed of light $c = f \lambda$,we have $\lambda = \frac{c}{f}$.
Substituting this into the momentum formula,we get $p = \frac{hf}{c}$.
Given: $h = 6.63 \times 10^{-34} \ J \ s$,$f = 1.5 \times 10^{13} \ Hz$,and $c = 3 \times 10^8 \ m/s$.
$p = \frac{6.63 \times 10^{-34} \times 1.5 \times 10^{13}}{3 \times 10^8}$.
$p = \frac{9.945 \times 10^{-21}}{3 \times 10^8} \approx 3.315 \times 10^{-29} \ kg \ m/s$.
Rounding to the given options,the correct value is $3.3 \times 10^{-29} \ kg \ m/s$.

(B) The energy $E$ and momentum $p$ of a photon are related to its wavelength $\lambda$ by the following equations:
$E = \frac{hc}{\lambda}$
$p = \frac{h}{\lambda}$
Where $h$ is Planck's constant and $c$ is the speed of light.
Since both $E$ and $p$ are inversely proportional to the wavelength $\lambda$ ($E \propto \frac{1}{\lambda}$ and $p \propto \frac{1}{\lambda}$),if the wavelength $\lambda$ is decreased,both the energy $E$ and the momentum $p$ will increase.

(C) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Using $hc = 12400 \ \text{eV} \cdot \mathring{A}$ and $\lambda = 4000 \ \mathring{A}$:
$E = \frac{12400}{4000} \ \text{eV} = 3.1 \ \text{eV}$.
To convert this energy into Joules,we multiply by the charge of an electron $(1.6 \times 10^{-19} \ \text{C})$:
$E = 3.1 \times 1.6 \times 10^{-19} \ \text{J} = 4.96 \times 10^{-19} \ \text{J}$.

(C) The relativistic mass $m$ of a particle is given by the formula: $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$,where $m_0$ is the rest mass,$v$ is the velocity,and $c$ is the speed of light.
Given that the mass becomes twice its rest mass,we have $m = 2m_0$.
Substituting this into the formula: $2m_0 = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$.
Dividing both sides by $m_0$: $2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$.
Squaring both sides: $4 = \frac{1}{1 - \frac{v^2}{c^2}}$.
Rearranging the terms: $1 - \frac{v^2}{c^2} = \frac{1}{4}$.
Thus,$\frac{v^2}{c^2} = 1 - \frac{1}{4} = \frac{3}{4}$.
Taking the square root: $v = \frac{\sqrt{3}}{2}c$.

(A) The momentum of a photon is given by $p = \frac{h}{\lambda}$.
Substituting the values: $p = \frac{6.63 \times 10^{-34} \ J \ s}{4400 \times 10^{-10} \ m} \approx 1.5 \times 10^{-27} \ kg \ m/s$.
The effective mass of a photon is given by $m = \frac{p}{c}$.
Substituting the values: $m = \frac{1.5 \times 10^{-27} \ kg \ m/s}{3 \times 10^8 \ m/s} = 5 \times 10^{-36} \ kg$.

(A) The visible light spectrum ranges from approximately $380 \, nm$ (violet) to $750 \, nm$ (red).
The maximum wavelength $(\lambda_{max})$ in the visible spectrum is $750 \, nm$.
The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 1240 \, eV \cdot nm$, we get:
$E = \frac{1240 \, eV \cdot nm}{750 \, nm} \approx 1.65 \, eV$.
Rounding to the nearest provided option, the value is $1.6 \, eV$.