Particle Nature of Light : Photon Questions in English

(A) The process of creating an electron-positron pair by a photon is called $pair \ production$. During this process,the energy of the photon is converted into the rest mass energy of the electron-positron pair and their respective kinetic energies.
The energy of the photon $(E)$ is given by:
$E = (m_e c^2 + K_e) + (m_p c^2 + K_p)$
Where $m_e c^2 = m_p c^2 = 0.51 \ MeV$ is the rest mass energy of an electron/positron,and $K_e = K_p = 0.29 \ MeV$ is the kinetic energy of each particle.
$E = (0.51 \ MeV + 0.29 \ MeV) + (0.51 \ MeV + 0.29 \ MeV)$
$E = 0.80 \ MeV + 0.80 \ MeV = 1.60 \ MeV$.

(B) Given: Wavelength $\lambda = 300 \ m$,Power $P = 10 \ kW = 10^4 \ W$.
Energy of one photon $E = \frac{hc}{\lambda}$.
The number of photons emitted per second $n = \frac{P}{E} = \frac{P \lambda}{hc}$.
Substituting the values: $n = \frac{10^4 \times 300}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{3 \times 10^6}{19.89 \times 10^{-26}} \approx 0.1508 \times 10^{32} = 1.508 \times 10^{31}$.
Thus,the number of photons emitted per second is approximately $1.5 \times 10^{31}$.

(C) The power $P$ of a light source is given by $P = n \cdot E$,where $n$ is the number of photons emitted per second and $E = \frac{hc}{\lambda}$ is the energy of a single photon.
Since the power $P$ is the same for both bulbs,we have $n_r E_r = n_b E_b$.
This implies $n_r \left( \frac{hc}{\lambda_r} \right) = n_b \left( \frac{hc}{\lambda_b} \right)$,which simplifies to $\frac{n_r}{\lambda_r} = \frac{n_b}{\lambda_b}$.
Since the wavelength of red light is greater than that of blue light $(\lambda_r > \lambda_b)$,it follows that $n_r > n_b$ to maintain the equality of power.
Therefore,the number of photons emitted by the red bulb is greater than that of the blue bulb.

(A) The intensity $I$ is given as $10^{-10} \ W/m^2$ and the area $A$ is $10^{-6} \ m^2$.
The power $P$ entering the eye is $P = I \times A = 10^{-10} \times 10^{-6} = 10^{-16} \ W$.
The energy of a single photon is $E = \frac{hc}{\lambda}$.
Given $\lambda = 560 \ nm = 560 \times 10^{-9} \ m$,$h = 6.6 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$:
$E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{560 \times 10^{-9}} = \frac{19.8 \times 10^{-26}}{560 \times 10^{-9}} \approx 3.54 \times 10^{-19} \ J$.
The number of photons per second $n$ is given by $n = \frac{P}{E}$.
$n = \frac{10^{-16}}{3.54 \times 10^{-19}} = \frac{1000}{3.54} \approx 282 \approx 3 \times 10^2$ photons.

(A) In a vacuum,all electromagnetic waves,including photons,travel at a constant speed $c$,where $c \approx 3 \times 10^8 \ m/s$.
Since the velocity $v = c$ is constant regardless of the frequency $f$,the graph of velocity versus frequency is a horizontal line.
This line is parallel to the frequency axis.
Therefore,the correct option is $A$.

(B) For an electron,the de Broglie wavelength is given by $\lambda = \frac{h}{p}$,which implies $p = \frac{h}{\lambda}$.
For a photon,the energy is given by $E = \frac{hc}{\lambda}$.
We are asked to find the ratio $p/E$.
Substituting the expressions for $p$ and $E$:
$\frac{p}{E} = \frac{h/\lambda}{hc/\lambda} = \frac{h}{\lambda} \times \frac{\lambda}{hc} = \frac{1}{c}$.
Given the speed of light $c = 3 \times 10^8 \ m/s$,we have:
$\frac{p}{E} = \frac{1}{3 \times 10^8} = 0.333 \times 10^{-8} = 3.33 \times 10^{-9} \ s/m$.

(C) The power $P$ of the transmitter is the total energy emitted per second.
Energy of one photon is given by $E = \frac{hc}{\lambda}$.
The number of photons $n$ emitted per second is given by $n = \frac{P}{E} = \frac{P\lambda}{hc}$.
Given: $P = 100 \ W$,$\lambda = 540 \times 10^{-9} \ m$,$h = 6 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$.
Substituting the values: $n = \frac{100 \times 540 \times 10^{-9}}{6 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{54000 \times 10^{-9}}{18 \times 10^{-26}} = \frac{5.4 \times 10^4 \times 10^{-9}}{1.8 \times 10^{-25}} = 3 \times 10^{20}$.
Thus,the number of photons emitted per second is $3 \times 10^{20}$.

(C) The intensity $I$ is given by $I = \frac{P}{A}$,where $P$ is the power and $A$ is the area.
Power $P = I \times A = \frac{nhc}{t\lambda}$,where $n$ is the number of photons,$h$ is Planck's constant,$c$ is the speed of light,and $t$ is time.
Rearranging for the number of photons per second $(n/t)$:
$\frac{n}{t} = \frac{IA\lambda}{hc}$
Substituting the given values:
$\frac{n}{t} = \frac{10^{-10} \times 10^{-6} \times 5.6 \times 10^{-7}}{6.63 \times 10^{-34} \times 3 \times 10^8}$
$\frac{n}{t} \approx \frac{5.6 \times 10^{-23}}{19.89 \times 10^{-26}} \approx 0.2815 \times 10^3 \approx 281.5$
Rounding to the nearest provided option,the number of photons is $300$.

(A) The energy of the photon is given as $E = 1 \, MeV = 1 \times 10^6 \, eV$.
Converting energy into Joules: $E = 10^6 \times 1.6 \times 10^{-19} \, J = 1.6 \times 10^{-13} \, J$.
The momentum $p$ of a photon is related to its energy $E$ by the formula $p = E/c$,where $c$ is the speed of light $(c \approx 3 \times 10^8 \, m/s)$.
Substituting the values: $p = \frac{1.6 \times 10^{-13} \, J}{3 \times 10^8 \, m/s}$.
$p \approx 0.533 \times 10^{-21} \, kg \, m/s$.
$p \approx 5.33 \times 10^{-22} \, kg \, m/s$.
Rounding to the nearest given option,the correct value is $5 \times 10^{-22} \, kg \, m/s$.

(C) The intensity $I$ of light from a point source at a distance $r$ is given by the inverse square law: $I \propto \frac{1}{r^2}$.
Since the number of photons emitted per unit area is proportional to the intensity,the number of photons $N$ incident on the photocell is proportional to the intensity.
Let $N_1$ be the number of photons at distance $r_1 = 0.5 \ m$ and $N_2$ be the number of photons at distance $r_2 = 1 \ m$.
Then,$\frac{N_1}{N_2} = \frac{r_2^2}{r_1^2}$.
Substituting the values: $\frac{N_1}{N_2} = \frac{1^2}{(0.5)^2} = \frac{1}{0.25} = 4$.
Therefore,the number of photons at $0.5 \ m$ is $4$ times the number of photons at $1 \ m$.

(D) The power $P$ of a monochromatic light beam is given by $P = N h \nu$,where $N$ is the number of photons emitted per second,$h$ is Planck's constant,and $\nu$ is the frequency.
Given:
Power $P = 2 \times 10^{-3} \text{ W}$
Frequency $\nu = 6.0 \times 10^{14} \text{ Hz}$
Planck's constant $h \approx 6.63 \times 10^{-34} \text{ J s}$
Energy of one photon $E = h \nu = (6.63 \times 10^{-34}) \times (6.0 \times 10^{14}) \text{ J} \approx 3.978 \times 10^{-19} \text{ J}$.
Number of photons emitted per second $N = P / E$:
$N = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}} \approx 0.5027 \times 10^{16} \approx 5 \times 10^{15}$ photons per second.
Thus,the correct option is $D$.

(A) Given: Wavelength $\lambda = 667 \, nm = 667 \times 10^{-9} \, m$. Power $P = 9 \, mW = 9 \times 10^{-3} \, W$.
Energy of a single photon $E = \frac{hc}{\lambda}$.
Using $h = 6.63 \times 10^{-34} \, J \cdot s$ and $c = 3 \times 10^8 \, m/s$:
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{667 \times 10^{-9}} \approx 2.98 \times 10^{-19} \, J$.
The number of photons per second $n = \frac{P}{E}$.
$n = \frac{9 \times 10^{-3}}{2.98 \times 10^{-19}} \approx 3.02 \times 10^{16} \approx 3 \times 10^{16}$ photons/sec.

(A) For source $S_1$:
Wavelength $\lambda_1 = 5000 \; \mathring{A}$,Number of photons per second $N_1 = 10^{15}$.
Power $P_1 = N_1 \times E_1 = N_1 \times \frac{hc}{\lambda_1}$.
For source $S_2$:
Wavelength $\lambda_2 = 5100 \; \mathring{A}$,Number of photons per second $N_2 = 1.02 \times 10^{15}$.
Power $P_2 = N_2 \times E_2 = N_2 \times \frac{hc}{\lambda_2}$.
Ratio $\frac{P_2}{P_1} = \frac{N_2 \cdot hc / \lambda_2}{N_1 \cdot hc / \lambda_1} = \frac{N_2}{N_1} \times \frac{\lambda_1}{\lambda_2}$.
Substituting the values:
$\frac{P_2}{P_1} = \frac{1.02 \times 10^{15}}{10^{15}} \times \frac{5000}{5100} = 1.02 \times \frac{50}{51} = \frac{1.02}{51} \times 50 = 0.02 \times 50 = 1$.
Thus,the ratio is $1$.

(C) The specific charge is defined as the ratio of charge to mass,denoted by $\frac{q}{m}$.
For an electron,$q = e$ and $m = m_e$.
For a proton,$q = e$ and $m = 1836 m_e$.
For an $\alpha$-particle,$q = 2e$ and $m = 4 m_p \approx 7344 m_e$.
For a $\beta$-particle,it is an electron,so $q = e$ and $m = m_e$.
Comparing the ratios:
Electron: $\frac{e}{m_e}$
Proton: $\frac{e}{1836 m_e}$
$\alpha$-particle: $\frac{2e}{7344 m_e} = \frac{e}{3672 m_e}$
$\beta$-particle: $\frac{e}{m_e}$
Since the denominator is largest for the $\alpha$-particle,the value of $\frac{q}{m}$ is the least for the $\alpha$-particle.

(B) The rate of photon emission (or incidence) is given by the formula $\frac{n}{t} = \frac{P}{E} = \frac{I \cdot A}{hc/\lambda} = \frac{I \cdot A \cdot \lambda}{hc}$.
Given:
Intensity $I = 150 \, mW/m^2 = 150 \times 10^{-3} \, W/m^2 = 0.15 \, W/m^2$.
Area $A = 4 \, cm^2 = 4 \times 10^{-4} \, m^2$.
Wavelength $\lambda = 3000 \, \mathring{A} = 3000 \times 10^{-10} \, m = 3 \times 10^{-7} \, m$.
Planck's constant $h = 6.63 \times 10^{-34} \, J \cdot s$.
Speed of light $c = 3 \times 10^8 \, m/s$.
Substituting the values:
$\frac{n}{t} = \frac{0.15 \times 4 \times 10^{-4} \times 3 \times 10^{-7}}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
$\frac{n}{t} = \frac{1.8 \times 10^{-10}}{19.89 \times 10^{-26}} \approx 0.0905 \times 10^{16} = 9.05 \times 10^{13} \, s^{-1}$.
Rounding to the nearest option, the rate is $9 \times 10^{13} \, s^{-1}$.

(B) When a stationary electron and a stationary positron annihilate,they produce two photons to conserve both energy and momentum.
Let the energy of each photon be $E = \frac{hc}{\lambda}$.
According to the law of conservation of energy:
$E_1 + E_2 = E_{electron} + E_{positron}$
$\frac{hc}{\lambda} + \frac{hc}{\lambda} = m_0 c^2 + m_0 c^2$
$\frac{2hc}{\lambda} = 2 m_0 c^2$
$\frac{hc}{\lambda} = m_0 c^2$
Solving for $\lambda$:
$\lambda = \frac{h}{m_0 c}$.

(C) The power $P$ of a bulb is given by the total energy emitted per unit time,which is $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$.
Since both bulbs have equal power $P$ and operate for the same time,the total energy emitted is the same.
Therefore,$n_r \cdot \frac{hc}{\lambda_r} = n_b \cdot \frac{hc}{\lambda_b}$.
This simplifies to $n_r \cdot \frac{1}{\lambda_r} = n_b \cdot \frac{1}{\lambda_b}$,or $n_r = n_b \cdot \frac{\lambda_r}{\lambda_b}$.
We know that the wavelength of red light is greater than the wavelength of blue light,i.e.,$\lambda_r > \lambda_b$.
Since $\frac{\lambda_r}{\lambda_b} > 1$,it follows that $n_r > n_b$.

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

(A) The energy $E$ of a photon with frequency $f$ is given by the Planck-Einstein relation: $E = hf$.
According to Einstein's mass-energy equivalence principle,the energy of a photon is also given by $E = mc^2$,where $m$ is the relativistic mass of the photon.
Equating the two expressions for energy: $mc^2 = hf$.
We know that the momentum $p$ of a particle is given by $p = mc$.
Rearranging the energy equation: $mc = \frac{hf}{c}$.
Therefore,the momentum of the photon is $p = \frac{hf}{c}$.

(A) Photons are quanta of electromagnetic radiation.
In a vacuum,all electromagnetic waves,including photons,travel at the speed of light,$c = 3 \times 10^8 \ m/s$.
The speed of light in a vacuum is a universal constant and is independent of the frequency $\nu$ or wavelength $\lambda$ of the radiation.
Therefore,the velocity of photons is independent of the frequency $\nu$.

(D) According to the law of conservation of charge, the total electric charge of an isolated system must remain constant. An electron has a charge of $-e$. If it were to convert completely into energy (photons), the final charge would be $0$. This would violate the conservation of charge. To convert an electron into energy, it must annihilate with its antiparticle, the positron (which has a charge of $+e$). The process is: $e^- + e^+ \rightarrow \text{energy}$. Therefore, the claim is impossible because it violates charge conservation.

(A) The total electrical power input is $P_{in} = 200\, W$.
Given the efficiency is $25\%$, the effective power output as light is $P_{out} = 0.25 \times 200\, W = 50\, W$.
The energy of a single photon is given by $E = \frac{hc}{\lambda}$.
If $n$ is the number of photons emitted per second, then the total power is $P_{out} = nE = n \frac{hc}{\lambda}$.
Rearranging for $n$, we get $n = \frac{P_{out} \lambda}{hc}$.
Substituting the values: $n = \frac{50 \times 0.6 \times 10^{-6}}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{30 \times 10^{-6}}{19.89 \times 10^{-26}} \approx 1.5 \times 10^{20}$ photons per second.

(A) Given: Wavelength $\lambda = 660\,nm = 660 \times 10^{-9}\,m$,Power $P = 0.5\,kW = 0.5 \times 10^3\,W$,Pulse width $t = 60\,ms = 60 \times 10^{-3}\,s$,Planck's constant $h = 6.62 \times 10^{-34}\,Js$,Speed of light $c = 3 \times 10^8\,m/s$.
The total energy $E$ of the pulse is given by $E = P \times t$.
The energy of a single photon is $E_{photon} = \frac{hc}{\lambda}$.
The number of photons $n$ is given by $n = \frac{E}{E_{photon}} = \frac{P \times t \times \lambda}{h \times c}$.
Substituting the values:
$n = \frac{0.5 \times 10^3 \times 60 \times 10^{-3} \times 660 \times 10^{-9}}{6.62 \times 10^{-34} \times 3 \times 10^8}$.
Approximating $6.62 \approx 6.6$:
$n = \frac{0.5 \times 60 \times 660 \times 10^{-9}}{6.6 \times 3 \times 10^{-26}} = \frac{30 \times 660 \times 10^{-9}}{19.8 \times 10^{-26}} \approx \frac{19800 \times 10^{-9}}{19.8 \times 10^{-26}} \approx 1000 \times 10^{17} = 10^{20}$.
Thus,the number of photons is $10^{20}$.

(B) Initial momentum of the photon is $P_i = \frac{h}{\lambda}$.
Final wavelength $\lambda' = \lambda + \Delta \lambda = \lambda + 3\lambda = 4\lambda$.
Final momentum of the photon is $P_f = \frac{h}{4\lambda} = \frac{P_i}{4}$. Let $P = P_f = \frac{h}{4\lambda}$,then $P_i = 4P$.
Using conservation of momentum: $\vec{P}_i = \vec{P}_f + \vec{P}_e$,where $\vec{P}_e$ is the momentum of the recoiled electron.
$\vec{P}_e = \vec{P}_i - \vec{P}_f$.
Taking the direction of incident photon as the $x$-axis:
$\vec{P}_i = 4P \hat{i}$.
$\vec{P}_f = P \cos 60^o \hat{i} + P \sin 60^o \hat{j} = P(\frac{1}{2}) \hat{i} + P(\frac{\sqrt{3}}{2}) \hat{j}$.
$\vec{P}_e = (4P - \frac{P}{2}) \hat{i} - \frac{\sqrt{3}P}{2} \hat{j} = \frac{7P}{2} \hat{i} - \frac{\sqrt{3}P}{2} \hat{j}$.
The angle $\phi$ of the recoiled electron is given by $\tan \phi = |\frac{P_{ey}}{P_{ex}}| = \frac{\sqrt{3}P/2}{7P/2} = \frac{\sqrt{3}}{7}$.
$\tan \phi = \frac{1.732}{7} \approx 0.247 \approx 0.25$.

(B) The power $P$ of the transmitter is given by $P = n \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.
Given:
Power $P = 10\, kW = 10^4\, W$
Wavelength $\lambda = 500\, m$
Planck's constant $h \approx 6.63 \times 10^{-34}\, J\cdot s$
Speed of light $c = 3 \times 10^8\, m/s$
Rearranging the formula for $n$:
$n = \frac{P \lambda}{hc}$
Substituting the values:
$n = \frac{10^4 \times 500}{6.63 \times 10^{-34} \times 3 \times 10^8}$
$n = \frac{5 \times 10^6}{19.89 \times 10^{-26}}$
$n \approx 0.251 \times 10^{32} \approx 2.51 \times 10^{31}$
Therefore,the order of magnitude is $10^{31}$.

(D) The power $P$ of the laser is given by $P = n \cdot E_{photon},$ where $n$ is the number of photons emitted per second and $E_{photon} = \frac{hc}{\lambda}.$
Given $P = 2\,mW = 2 \times 10^{-3}\,W,$ $\lambda = 500\,nm = 500 \times 10^{-9}\,m,$ $h = 6.6 \times 10^{-34}\,Js,$ and $c = 3.0 \times 10^8\,m/s.$
Rearranging for $n$: $n = \frac{P \cdot \lambda}{h \cdot c}$
Substituting the values:
$n = \frac{2 \times 10^{-3} \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3.0 \times 10^8}$
$n = \frac{1000 \times 10^{-12}}{19.8 \times 10^{-26}}$
$n = \frac{10^{-9}}{19.8 \times 10^{-26}} \approx 0.0505 \times 10^{17} = 5.05 \times 10^{15} \approx 5 \times 10^{15}.$
Thus,the number of photons emitted per second is $5 \times 10^{15}.$

(A) The total energy $E$ emitted by the bulb is given by $E = P \times t,$ where $P = 10\,W$ and $t = 10\,s.$
$E = 10 \times 10 = 100\,J.$
The energy of a single photon is given by $E_{ph} = \frac{hc}{\lambda}.$
Given $h = 6.63 \times 10^{-34}\,J\cdot s,$ $c = 3 \times 10^8\,m/s,$ and $\lambda = 1000 \times 10^{-10}\,m = 10^{-7}\,m.$
$E_{ph} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{10^{-7}} = 19.89 \times 10^{-19}\,J.$
The number of photons $n$ is given by $n = \frac{E}{E_{ph}}.$
$n = \frac{100}{19.89 \times 10^{-19}} \approx 5.027 \times 10^{20}.$
Rounding to the nearest provided option,the correct value is $5.05 \times 10^{20}.$

(B) Planck's constant $(h)$ relates the energy of a photon to its frequency according to the equation $E = h \nu$.
Since $E$ is measured in Joules $(J)$ and frequency $\nu$ is in $s^{-1}$,the unit of $h$ is $J \cdot s$.
Alternatively,since $E = \text{Force} \times \text{Distance} = (kg \cdot m/s^2) \times m = kg \cdot m^2/s^2$,then $h = E/\nu = (kg \cdot m^2/s^2) / (1/s) = kg \cdot m^2/s$.
The numerical value of Planck's constant is approximately $6.63 \times 10^{-34} \; J \cdot s$ or $6.63 \times 10^{-34} \; kg \cdot m^2/s$.
Comparing this with the given options,option $B$ represents the correct value and units.

(B) The power $P$ of the light source is given by $P = n \cdot E_{photon}$,where $n$ is the number of photons emitted per second and $E_{photon} = \frac{hc}{\lambda}$.
Thus,$n = \frac{P \cdot \lambda}{hc}$.
Given: $P = 100\,W$,$\lambda = 5000 \times 10^{-10}\,m$,$h = 6.63 \times 10^{-34}\,J\cdot s$,and $c = 3 \times 10^8\,m/s$.
Substituting the values:
$n = \frac{100 \times 5000 \times 10^{-10}}{6.63 \times 10^{-34} \times 3 \times 10^8}$
$n = \frac{5 \times 10^{-5}}{19.89 \times 10^{-26}} \approx 0.25 \times 10^{21} = 2.5 \times 10^{20}$.
Therefore,the rate of emission of photons is of the order of $10^{20}$.

(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Here,$h = 6.6 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^{8} \, m/s$,and $\lambda = 2 \, cm = 2 \times 10^{-2} \, m$.
Substituting the values:
$E = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2 \times 10^{-2}} = 9.9 \times 10^{-24} \, J$.
To convert this energy into $eV$,divide by the charge of an electron $(1.6 \times 10^{-19} \, C)$:
$E = \frac{9.9 \times 10^{-24}}{1.6 \times 10^{-19}} \, eV = 6.1875 \times 10^{-5} \, eV \approx 6.2 \times 10^{-5} \, eV$.

(B) The energy of a single photon is given by $E_p = \frac{hc}{\lambda}$.
Given: $h = 6.63 \times 10^{-34} \, \text{J s}$,$c = 3 \times 10^8 \, \text{m/s}$,and $\lambda = 5000 \, \mathring{A} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m}$.
$E_p = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{5 \times 10^{-7}} \approx 3.978 \times 10^{-19} \, \text{J} \approx 4 \times 10^{-19} \, \text{J}$.
Let $n$ be the number of photons. The total energy $E_{total} = n \times E_p$.
Given $E_{total} = 10^{-7} \, \text{J}$.
$n = \frac{E_{total}}{E_p} = \frac{10^{-7}}{4 \times 10^{-19}} = 0.25 \times 10^{12} = 2.5 \times 10^{11}$.
Thus,the number of photons is $2.5 \times 10^{11}$.

(A) The power supplied to the bulb is $P = 100 \, W = 100 \, J/s$.
Energy radiated as visible light per second is $E = 5\% \text{ of } 100 \, J/s = \frac{5}{100} \times 100 = 5 \, J/s$.
The energy of a single photon is given by $E_{photon} = \frac{hc}{\lambda}$.
Given $\lambda = 5.6 \times 10^{-5} \, cm = 5.6 \times 10^{-7} \, m$,$h = 6.6 \times 10^{-34} \, J \cdot s$,and $c = 3 \times 10^8 \, m/s$.
Let $n$ be the number of photons emitted per second.
Then $n \times E_{photon} = 5 \, J/s$.
$n = \frac{5 \times \lambda}{hc} = \frac{5 \times 5.6 \times 10^{-7}}{(6.6 \times 10^{-34}) \times (3 \times 10^8)}$.
$n = \frac{28 \times 10^{-7}}{19.8 \times 10^{-26}} \approx 1.414 \times 10^{19} \approx 1.4 \times 10^{19}$ photons per second.

(A) The total energy emitted by the lamp in time $t$ is $E_{total} = P \times t$.
Since the lamp radiates light uniformly in all directions,the intensity at a distance $l$ is $I = \frac{P}{4\pi l^2}$.
The sensor has a circular opening of diameter $4d$,so its radius is $r = 2d$.
The area of the circular opening is $A = \pi r^2 = \pi (2d)^2 = 4\pi d^2$.
The energy $E$ incident on the sensor is $E = I \times A \times t = \left( \frac{P}{4\pi l^2} \right) \times (4\pi d^2) \times t = \frac{P d^2 t}{l^2}$.
If $n$ is the number of photons,then $E = n \times \frac{hc}{\lambda}$.
Equating the two expressions for energy: $n \frac{hc}{\lambda} = \frac{P d^2 t}{l^2}$.
Solving for $n$: $n = \frac{P \lambda d^2 t}{hc l^2}$.

(C) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Using the shortcut formula $E(\text{eV}) = \frac{12400}{\lambda(\mathring{A})}$,we get:
$E = \frac{12400}{1000} \, \text{eV} = 12.4 \, \text{eV}$.
To convert this energy into Joules,we multiply by the charge of an electron $(1.6 \times 10^{-19} \, \text{C})$:
$E = 12.4 \times 1.6 \times 10^{-19} \, \text{J}$.
$E = 19.84 \times 10^{-19} \, \text{J}$.

(B) The energy of a photon is given by $E = pc$,where $p$ is the momentum and $c$ is the speed of light.
Thus,the momentum $p = E/c$.
Given $E = 20 \, eV = 20 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-18} \, J$.
The speed of light $c = 3 \times 10^8 \, m/s$.
Substituting these values:
$p = \frac{3.2 \times 10^{-18}}{3 \times 10^8} \, kg \cdot m/s$.
$p = 1.066 \times 10^{-26} \, kg \cdot m/s = 10.66 \times 10^{-27} \, kg \cdot m/s$.
Therefore,the correct option is $B$.

(B) The energy $E$ of a photon is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
The momentum $P$ of a photon is given by $P = \frac{h}{\lambda}$.
From these relations,it is clear that both $E$ and $P$ are inversely proportional to the wavelength $\lambda$ ($E \propto \frac{1}{\lambda}$ and $P \propto \frac{1}{\lambda}$).
Therefore,if the wavelength $\lambda$ is reduced,both the energy $E$ and the momentum $P$ will increase.

(B) The momentum of a photon is given by $p = \frac{h}{\lambda}$. Since $p = mc$ (where $m$ is the relativistic mass of the photon and $c$ is the speed of light), we have $mc = \frac{h}{\lambda}$, which implies $m = \frac{h}{c\lambda}$. Thus, the mass $m$ of a moving photon is inversely proportional to its wavelength $\lambda$. The Assertion is correct.
The Reason states $E = mc^2$. While this is Einstein's mass-energy equivalence relation, it describes the energy of a particle at rest or the energy equivalent of a mass. It does not explain the relationship between the mass of a photon and its wavelength. Therefore, the Reason is not the correct explanation for the Assertion.

(A) According to the theory of relativity,the energy of a particle is given by $E^2 = (pc)^2 + (m_0c^2)^2$.
For a photon,the rest mass $m_0$ is $0$.
Therefore,$E^2 = (pc)^2$,which simplifies to $E = pc$ or $p = E/c$.
This relationship is derived specifically from the particle nature of the photon,which allows it to possess momentum despite having no rest mass.
Thus,the Assertion is correct,and the Reason is the correct explanation for the Assertion.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given:
$h = 6.63 \times 10^{-34} \, J \cdot s$
$c = 3 \times 10^8 \, m/s$
$\lambda = 6840 \times 10^{-10} \, m$
Using the relation $E(eV) = \frac{12400}{\lambda(\mathring{A})}$,we get:
$E = \frac{12400}{6840} \, eV$
$E \approx 1.81 \, eV$.

(N/A) The electron is faster. The ratio of their speeds is $13.54:1$.
Given:
Mass of the electron,$m_e = 9.11 \times 10^{-31} \; kg$
Mass of the proton,$m_p = 1.67 \times 10^{-27} \; kg$
Kinetic energy of the electron,$E_{Ke} = 10 \; keV = 10^4 \; eV = 1.60 \times 10^{-15} \; J$
Kinetic energy of the proton,$E_{Kp} = 100 \; keV = 10^5 \; eV = 1.60 \times 10^{-14} \; J$
Using the formula for kinetic energy $E_K = \frac{1}{2}mv^2$,the velocity is $v = \sqrt{\frac{2E_K}{m}}$.
For the electron:
$v_e = \sqrt{\frac{2 \times 1.60 \times 10^{-15}}{9.11 \times 10^{-31}}} \approx 5.93 \times 10^7 \; m/s$
For the proton:
$v_p = \sqrt{\frac{2 \times 1.60 \times 10^{-14}}{1.67 \times 10^{-27}}} \approx 4.38 \times 10^6 \; m/s$
Comparing the velocities,$v_e > v_p$,so the electron is faster.
The ratio of their speeds is $\frac{v_e}{v_p} = \frac{5.93 \times 10^7}{4.38 \times 10^6} \approx 13.54$.

(A) The concept of 'ether' was proposed as a medium for the propagation of light waves. However,the Michelson-Morley experiment failed to detect the existence of ether. Albert Einstein,in his theory of special relativity published in $1905$,discarded the need for an ether medium,stating that the speed of light in a vacuum is constant for all inertial observers,regardless of the motion of the source or the observer.

(N/A) Reflection and refraction arise through the interaction of incident light with the atomic constituents of matter. Atoms act as oscillators that take up the frequency of the external agency (incident light),resulting in forced oscillations. Since the frequency of light emitted by a charged oscillator equals its frequency of oscillation,the frequency of reflected and refracted light remains equal to the incident frequency.
$(b)$ No. The energy carried by a wave depends on the amplitude of the wave,not on the speed of wave propagation. Therefore,a reduction in speed does not imply a reduction in energy.
$(c)$ In the photon picture of light,the intensity is determined by the number of photons crossing a unit area per unit time for a given frequency.

(N/A) The energy of each photon is given by $E = h\nu$,where $h = 6.63 \times 10^{-34} \;J \cdot s$ and $\nu = 6.0 \times 10^{14} \;Hz$.
$E = (6.63 \times 10^{-34} \;J \cdot s)(6.0 \times 10^{14} \;Hz) = 3.978 \times 10^{-19} \;J \approx 3.98 \times 10^{-19} \;J$.
$(b)$ If $N$ is the number of photons emitted per second,the total power $P$ is given by $P = N \cdot E$.
Therefore,$N = \frac{P}{E} = \frac{2.0 \times 10^{-3} \;W}{3.978 \times 10^{-19} \;J} \approx 5.03 \times 10^{15} \;\text{photons/s}$.

(N/A) Given: Wavelength $\lambda = 632.8 \; nm = 632.8 \times 10^{-9} \; m$,Power $P = 9.42 \; mW = 9.42 \times 10^{-3} \; W$.
$(a)$ Energy of a photon $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} \; J \cdot s)(3 \times 10^8 \; m/s)}{632.8 \times 10^{-9} \; m} \approx 3.14 \times 10^{-19} \; J$.
Momentum of a photon $p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34} \; J \cdot s}{632.8 \times 10^{-9} \; m} \approx 1.05 \times 10^{-27} \; kg \cdot m/s$.
$(b)$ Number of photons per second $N = \frac{P}{E} = \frac{9.42 \times 10^{-3} \; W}{3.14 \times 10^{-19} \; J} = 3 \times 10^{16} \; \text{photons/second}$.
$(c)$ Momentum of a hydrogen atom $p = mv$. Given $p = 1.05 \times 10^{-27} \; kg \cdot m/s$ and mass of hydrogen atom $m \approx 1.67 \times 10^{-27} \; kg$.
$v = \frac{p}{m} = \frac{1.05 \times 10^{-27}}{1.67 \times 10^{-27}} \approx 0.63 \; m/s$.

(D) Given: Energy flux $I = 1.388 \times 10^{3} \; W/m^{2}$, Wavelength $\lambda = 550 \; nm = 550 \times 10^{-9} \; m$.
The energy of a single photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \; J \cdot s$ and $c = 3 \times 10^{8} \; m/s$:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{550 \times 10^{-9}} \approx 3.614 \times 10^{-19} \; J$.
The number of photons $n$ incident per unit area per second is given by $n = \frac{I}{E}$.
$n = \frac{1.388 \times 10^{3}}{3.614 \times 10^{-19}} \approx 3.84 \times 10^{21} \; \text{photons}/m^{2} \cdot s$.
Rounding to the nearest value, we get $n \approx 4 \times 10^{21} \; \text{photons}/m^{2} \cdot s$.

(N/A) Given: Power $P = 100 \; W$, Wavelength $\lambda = 589 \; nm = 589 \times 10^{-9} \; m$, Planck's constant $h = 6.63 \times 10^{-34} \; J \cdot s$, Speed of light $c = 3 \times 10^8 \; m/s$.
$(a)$ The energy per photon $E$ is given by the formula $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{(6.63 \times 10^{-34} \; J \cdot s) \times (3 \times 10^8 \; m/s)}{589 \times 10^{-9} \; m} \approx 3.376 \times 10^{-19} \; J$.
$(b)$ The rate of delivery of photons $n$ is given by $n = \frac{P}{E}$.
Substituting the values: $n = \frac{100 \; W}{3.376 \times 10^{-19} \; J} \approx 2.96 \times 10^{20} \; \text{photons/second}$.

(N/A) The momentum $p$ of a photon is given by the relation $p = E / c$,where $E$ is the energy of the photon and $c$ is the speed of light.
Since the energy of a photon is $E = h \nu$,where $h$ is Planck's constant and $\nu$ is the frequency,we have $p = h \nu / c$.
We know that the speed of light is related to frequency and wavelength by $c = \nu \lambda$,which implies $\lambda = c / \nu$.
Substituting $c / \nu = \lambda$ into the momentum equation,we get $p = h / \lambda$.
Rearranging this gives the de Broglie wavelength formula: $\lambda = h / p$.
Thus,the de Broglie wavelength of a photon is exactly equal to the wavelength of the electromagnetic radiation of which the photon is the quantum.

(C) Total energy of the electron-positron pair is $E = 10.2 \; BeV = 10.2 \times 10^9 \; eV$.
Converting this energy into Joules: $E = 10.2 \times 10^9 \times 1.602 \times 10^{-19} \; J = 1.634 \times 10^{-9} \; J$.
Since the pair annihilates into two $\gamma$-rays of equal energy,the energy of each $\gamma$-ray is $E' = E / 2 = 0.817 \times 10^{-9} \; J$.
Using the relation $E' = \frac{hc}{\lambda}$,the wavelength $\lambda$ is given by $\lambda = \frac{hc}{E'}$.
Substituting $h = 6.626 \times 10^{-34} \; Js$ and $c = 3 \times 10^8 \; m/s$:
$\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.817 \times 10^{-9}} \approx 2.433 \times 10^{-16} \; m$.
Rounding to the provided option,the correct wavelength is $2.436 \times 10^{-16} \; m$.

(N/A) Power of the medium wave transmitter,$P = 10\; kW = 10^4\; W = 10^4\; J/s$.
Hence,energy emitted by the transmitter per second,$E = 10^4\; J$.
Wavelength of the radio wave,$\lambda = 500\; m$.
The energy of a single photon is given by $E_1 = \frac{hc}{\lambda}$.
Where,$h = 6.6 \times 10^{-34}\; Js$ and $c = 3 \times 10^8\; m/s$.
$E_1 = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{500} = 3.96 \times 10^{-28}\; J$.
Let $n$ be the number of photons emitted per second.
$n = \frac{E}{E_1} = \frac{10^4}{3.96 \times 10^{-28}} \approx 2.525 \times 10^{31} \approx 3 \times 10^{31}$ photons/s.
Since the number of photons is extremely large,the discrete nature of energy can be ignored,and the wave can be treated as continuous.
$(b)$ Intensity of light,$I = 10^{-10}\; W m^{-2}$.
Area of the pupil,$A = 0.4\; cm^2 = 0.4 \times 10^{-4}\; m^2$.
Frequency of white light,$\nu = 6 \times 10^{14}\; Hz$.
Energy of one photon,$E_p = h\nu = 6.6 \times 10^{-34} \times 6 \times 10^{14} = 3.96 \times 10^{-19}\; J$.
Total energy incident on the pupil per second,$E_{total} = I \times A = 10^{-10} \times 0.4 \times 10^{-4} = 4 \times 10^{-15}\; J/s$.
Number of photons entering the pupil per second,$n = \frac{E_{total}}{E_p} = \frac{4 \times 10^{-15}}{3.96 \times 10^{-19}} \approx 1.01 \times 10^4$ photons/s.
This number is large enough that the human eye cannot perceive individual photons.

(N/A) The unit of energy in atomic and nuclear physics is the electron volt $(eV)$.
Definition: One electron volt $(1 eV)$ is defined as the amount of kinetic energy gained by a single electron when it is accelerated through a potential difference of $1$ volt.
Mathematically,the relationship between electron volt and Joule is:
$1 eV = 1.602 \times 10^{-19} J$
$1 J = 6.242 \times 10^{18} eV$
Table $11.1$: Work function of some metals

Metal	Work function $\phi_{0} (eV)$
$Cs$	$2.14$
$K$	$2.30$
$Na$	$2.75$
$Ca$	$3.20$
$Mo$	$4.17$
$Pb$	$4.25$
$Al$	$4.28$
$Hg$	$4.49$
$Cu$	$4.65$
$Ag$	$4.70$
$Ni$	$5.15$
$Pt$	$5.65$