Particle Nature of Light : Photon Questions in English

(A) Planck's constant is a fundamental physical constant denoted by the symbol $h$.
It relates the energy of a photon to its frequency,given by the equation $E = h\nu$.
The accepted value of Planck's constant is approximately $6.626 \times 10^{-34} \, J \cdot s$.
Rounding this to two decimal places gives $6.63 \times 10^{-34} \, J \cdot s$.
Therefore,the correct option is $A$.

(C) According to the theory of special relativity,the length of an object moving at relativistic speeds undergoes length contraction in the direction of motion.
The formula for length contraction is $l = l_0 \sqrt{1 - \frac{v^2}{c^2}}$,where $l_0$ is the proper length,$v$ is the velocity of the object,and $c$ is the speed of light.
Given: $l_0 = 1 \, m$,$v = 2.7 \times 10^8 \, m/s$,and $c = 3 \times 10^8 \, m/s$.
Substituting the values into the formula:
$l = 1 \times \sqrt{1 - \left( \frac{2.7 \times 10^8}{3 \times 10^8} \right)^2}$
$l = \sqrt{1 - (0.9)^2}$
$l = \sqrt{1 - 0.81}$
$l = \sqrt{0.19}$
$l \approx 0.4358 \, m \approx 0.44 \, m$.
Thus,the apparent length of the stick is $0.44 \, m$.

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.
For a particle with mass $m$ and kinetic energy $E$,the momentum is $p = \sqrt{2mE}$. Thus,$\lambda = \frac{h}{\sqrt{2mE}}$.
Rearranging for energy,we get $E = \frac{h^2}{2m\lambda^2}$.
Since the wavelength $\lambda$ is the same for all,the energy $E$ is inversely proportional to the mass $m$ $(E \propto \frac{1}{m})$.
For a photon,the energy is given by $E = \frac{hc}{\lambda}$. Comparing this to the massive particles,the photon behaves as if it has the smallest effective mass in this context,leading to the highest energy for a given wavelength.
Therefore,the photon has the most energy.

(A) According to the theory of relativity,the energy $E$ and momentum $p$ of a particle are related to its rest mass $m_0$ by the equation $E^2 = (pc)^2 + (m_0c^2)^2$.
For a particle with zero rest mass $(m_0 = 0)$,the equation simplifies to $E = pc$.
Since $p = mv$ and $E = mc^2$ (where $m$ is the relativistic mass),we have $mc^2 = mvc$,which implies $v = c$.
Therefore,any particle with zero rest mass,such as a photon,must travel at the speed of light $c$ in a vacuum.

(D) The momentum $p$ of a photon is related to its frequency $\nu$ by the formula: $p = \frac{h\nu}{c}$.
Rearranging for frequency $\nu$,we get: $\nu = \frac{pc}{h}$.
Given: $p = 3.3 \times 10^{-29} \ kg \ m/s$,$c = 3 \times 10^8 \ m/s$,and Planck's constant $h = 6.6 \times 10^{-34} \ J \ s$.
Substituting the values: $\nu = \frac{3.3 \times 10^{-29} \times 3 \times 10^8}{6.6 \times 10^{-34}}$.
$\nu = \frac{9.9 \times 10^{-21}}{6.6 \times 10^{-34}} = 1.5 \times 10^{13} \ Hz$.

(D) The energy of a photon is directly proportional to its frequency,with the proportionality constant being Planck's constant $h$.
The relationship between energy $E$,frequency $\nu$,and wavelength $\lambda$ is given by $E = h\nu$.
Since the speed of light $c$ is related to frequency and wavelength by $c = \nu\lambda$,we can express frequency as $\nu = c/\lambda$.
Substituting this into the energy equation,we get $E = h(c/\lambda) = hc/\lambda$.

(C) The relationship between the energy $E$ and momentum $p$ of a photon is given by $E = p \times c$,where $c$ is the speed of light.
Given momentum $p = 2 \times 10^{-16} \text{ g} \cdot \text{cm/s}$.
The speed of light $c = 3 \times 10^{10} \text{ cm/s}$.
Substituting these values into the formula:
$E = (2 \times 10^{-16} \text{ g} \cdot \text{cm/s}) \times (3 \times 10^{10} \text{ cm/s})$
$E = 6 \times 10^{-6} \text{ g} \cdot \text{cm}^2/\text{s}^2$
Since $1 \text{ erg} = 1 \text{ g} \cdot \text{cm}^2/\text{s}^2$,the energy is $6 \times 10^{-6} \text{ erg}$.

(A) According to Einstein's quantum theory,light propagates in the form of packets,i.e.,quanta of energy,which are called photons.
The rest mass of a photon is zero. This can be shown using the theory of relativity.
According to the relativistic mass equation,the mass of a particle is given by:
$m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$
Rearranging for the rest mass $(m_{0})$:
$m_{0} = m \sqrt{1 - \frac{v^{2}}{c^{2}}}$
Since a photon always travels at the speed of light $(v = c)$,we substitute $v = c$ into the equation:
$m_{0} = m \sqrt{1 - \frac{c^{2}}{c^{2}}} = m \sqrt{1 - 1} = 0$
Therefore,the rest mass of a photon is $0$.

(A) The momentum $p$ of a photon is given by the de Broglie relation: $p = \frac{h}{\lambda}$.
Given:
Planck's constant $h = 6.6 \times 10^{-34}\,J\cdot s$
Wavelength $\lambda = 5000\,\mathring{A} = 5000 \times 10^{-10}\,m = 5 \times 10^{-7}\,m$.
Substituting the values:
$p = \frac{6.6 \times 10^{-34}}{5 \times 10^{-7}}$
$p = 1.32 \times 10^{-27}\,kg\cdot m/s$.
Thus,the correct option is $A$.

(B) According to Einstein's mass-energy equivalence relation,the energy of a photon is given by $E = mc^2$.
Also,the energy of a photon is given by $E = h\nu$.
Equating the two,we get $mc^2 = h\nu$.
Since momentum $p$ is defined as $p = mc$,we can rewrite the energy equation as $p \cdot c = h\nu$.
Therefore,the momentum of the photon is $p = \frac{h\nu}{c}$.

(D) According to Einstein's mass-energy equivalence principle,the energy $E$ of a photon is given by $E = h\nu$.
Also,from the mass-energy relation,$E = mc^2$,where $m$ is the relativistic mass of the photon.
Equating the two expressions for energy:
$mc^2 = h\nu$
Therefore,the mass of the photon in motion is $m = \frac{h\nu}{c^2}$.

(B) The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
From the relation between energy and momentum for a photon,$E = pc$,where $c$ is the speed of light.
Equating the two expressions for energy: $h\nu = pc$.
Therefore,the frequency is $\nu = \frac{pc}{h}$.

(B) The power $P$ of the transmitter is given by the total energy radiated per unit time,$P = \frac{E}{t}$.
Since the energy of a single photon is $E_{photon} = \frac{hc}{\lambda}$,the total energy radiated by $n$ photons in time $t$ is $E = n \times \frac{hc}{\lambda}$.
Therefore,the number of photons radiated per second is $\frac{n}{t} = \frac{P \lambda}{hc}$.
Given: $P = 10 \ kW = 10^4 \ W$,$\lambda = 300 \ m$,$h = 6.6 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$.
Substituting the values:
$\frac{n}{t} = \frac{10^4 \times 300}{6.6 \times 10^{-34} \times 3 \times 10^8}$
$\frac{n}{t} = \frac{3 \times 10^6}{19.8 \times 10^{-26}}$
$\frac{n}{t} = \frac{3}{19.8} \times 10^{32} \approx 0.1515 \times 10^{32} = 1.515 \times 10^{31} \approx 1.5 \times 10^{31}$ photons per second.

(A) The energy of a photon is given by $E = h\nu$ and the momentum is given by $p = \frac{h}{\lambda}$.
Since the velocity of a photon is $c = \nu \lambda$,we can express the energy as $E = h \left( \frac{c}{\lambda} \right) = \frac{hc}{\lambda}$.
Substituting the expression for momentum $p = \frac{h}{\lambda}$ into the energy equation,we get $E = pc$.
Rearranging this equation to solve for the velocity $c$,we obtain $c = \frac{E}{p}$.
Thus,the velocity of the photon is $E/p$.

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
For energy in $eV$ and wavelength in $\mathring A$,the relation is approximately $E(eV) = \frac{12400}{\lambda(\mathring A)}$.
Given $E = 2.48 \,eV$,we have $2.48 = \frac{12400}{\lambda}$.
$\lambda = \frac{12400}{2.48} \, \mathring A$.
$\lambda = 5000 \, \mathring A$.
Thus,the approximate wavelength is $5000 \, \mathring A$.

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given:
$h = 6.62 \times 10^{-34} \ J \cdot s$
$c = 3 \times 10^8 \ m/s$
$\lambda = 21 \ cm = 0.21 \ m$
To convert the energy from Joules to electron-volts $(eV)$,we divide by $1.6 \times 10^{-19} \ J/eV$.
$E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{0.21 \times 1.6 \times 10^{-19}} \ eV$
$E = \frac{19.86 \times 10^{-26}}{0.336 \times 10^{-19}} \ eV$
$E \approx 5.91 \times 10^{-6} \ eV$.
Therefore,the correct option is $B$.

(B) The momentum $p$ of a photon is given by the formula $p = \frac{h}{\lambda}$,where $h$ is Planck's constant and $\lambda$ is the wavelength.
Given: $h = 6.6 \times 10^{-34} \ J \ s$ and $\lambda = 10^{-10} \ m$.
Substituting the values:
$p = \frac{6.6 \times 10^{-34}}{10^{-10}}$
$p = 6.6 \times 10^{-24} \ kg \ m/s$.
Therefore,the correct option is $B$.

(A) The energy $E$ of a quantum (photon) is given by the formula $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Given:
Frequency $\nu = 10^{15} \ Hz$
Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$
Substituting the values into the formula:
$E = (6.6 \times 10^{-34} \ J \cdot s) \times (10^{15} \ Hz)$
$E = 6.6 \times 10^{-34+15} \ J$
$E = 6.6 \times 10^{-19} \ J$
Therefore,the correct option is $A$.

(A) The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
According to Einstein's mass-energy equivalence,$E = mc^2$,where $m$ is the relativistic mass and $c$ is the speed of light.
Equating the two expressions for energy: $mc^2 = h\nu$.
Since the momentum $p$ of a photon is defined as $p = mc$,we can substitute $m = \frac{p}{c}$ into the equation: $p = \frac{h\nu}{c}$.
Using the relation $\nu = \frac{c}{\lambda}$,we substitute for $\nu$: $p = \frac{h(c/\lambda)}{c} = \frac{h}{\lambda}$.
Therefore,the momentum of a photon is $\frac{h}{\lambda}$.

(B) The energy of a photon is given by the relation $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Given energy $E = 1 \;MeV = 1 \times 10^6 \;eV = 10^6 \times 1.6 \times 10^{-19} \;J = 1.6 \times 10^{-13} \;J$.
Planck's constant $h \approx 6.63 \times 10^{-34} \;J \cdot s$.
Using the formula $\nu = \frac{E}{h}$:
$\nu = \frac{1.6 \times 10^{-13} \;J}{6.63 \times 10^{-34} \;J \cdot s} \approx 0.241 \times 10^{21} \;Hz = 2.41 \times 10^{20} \;Hz$.
Thus,the frequency is approximately $2.4 \times 10^{20} \;Hz$.

(A) The momentum $p$ of a photon is given by the formula $p = \frac{E}{c}$,where $E = h\nu$ is the energy of the photon.
Substituting the values,we get $p = \frac{h\nu}{c}$.
Given: $h = 6.6 \times 10^{-34} \ J \ s$,$\nu = 1.5 \times 10^{13} \ Hz$,and $c = 3 \times 10^8 \ m/s$.
$p = \frac{6.6 \times 10^{-34} \times 1.5 \times 10^{13}}{3 \times 10^8}$.
$p = \frac{9.9 \times 10^{-21}}{3 \times 10^8} = 3.3 \times 10^{-29} \ kg \ m/s$.

(A) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given:
Planck's constant $h = 6.62 \times 10^{-34} \ J \cdot s$
Speed of light $c = 3 \times 10^8 \ m/s$
Wavelength $\lambda = 450 \ nm = 450 \times 10^{-9} \ m$
Substituting these values into the formula:
$E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{450 \times 10^{-9}}$
$E = \frac{19.86 \times 10^{-26}}{450 \times 10^{-9}}$
$E = 0.04413 \times 10^{-17} \ J = 4.413 \times 10^{-19} \ J$
Rounding to two significant figures,we get $E \approx 4.4 \times 10^{-19} \ J$.

(C) The energy of a photon is given by the relation $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Given energy $E = 66 \ eV = 66 \times 1.6 \times 10^{-19} \ J$.
Planck's constant $h \approx 6.6 \times 10^{-34} \ J \cdot s$.
Substituting the values into the formula: $\nu = \frac{E}{h}$.
$\nu = \frac{66 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}$.
$\nu = 10 \times 1.6 \times 10^{15} \ Hz$.
$\nu = 16 \times 10^{15} \ Hz$.

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
To express $E$ in $eV$ and $\lambda$ in $\mathring{A}$,we use:
$E(eV) = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{(1.602 \times 10^{-19} \text{ J/eV}) \times (\lambda \times 10^{-10} \text{ m})}$
$E(eV) \approx \frac{12400}{\lambda(\mathring{A})}$.
To convert this to $KeV$,we divide by $1000$:
$E(KeV) = \frac{12400}{1000 \times \lambda(\mathring{A})} = \frac{12.4}{\lambda(\mathring{A})}$.
Thus,the correct relation is $E = 12.4/\lambda$.

(B) The energy of a photon is given by the formula $E = h\nu$,where $E$ is the energy,$h$ is Planck's constant,and $\nu$ is the frequency.
First,convert the energy from electron-volts $(eV)$ to Joules $(J)$:
$E = 100 \ eV = 100 \times 1.6 \times 10^{-19} \ J = 1.6 \times 10^{-17} \ J$.
Now,substitute the values into the formula:
$1.6 \times 10^{-17} = 6.6 \times 10^{-34} \times \nu$.
Solving for $\nu$:
$\nu = \frac{1.6 \times 10^{-17}}{6.6 \times 10^{-34}} \approx 0.2424 \times 10^{17} \ Hz = 2.42 \times 10^{16} \ Hz$.
Thus,the correct option is $B$.

(A) The momentum $p$ of a photon is given by the formula $p = \frac{h}{\lambda}$.
Given $h = 6.6 \times 10^{-34} \, J \cdot s$ and $\lambda = 4400 \times 10^{-10} \, m$.
$p = \frac{6.6 \times 10^{-34}}{4400 \times 10^{-10}} = 1.5 \times 10^{-27} \, kg \cdot m/s$.
The effective mass $m$ of a photon is given by $m = \frac{p}{c}$,where $c = 3 \times 10^8 \, m/s$.
$m = \frac{1.5 \times 10^{-27}}{3 \times 10^8} = 5 \times 10^{-36} \, kg$.
Thus,the effective mass is $5 \times 10^{-36} \, kg$ and the momentum is $1.5 \times 10^{-27} \, kg \cdot m/s$.

(A) photon is a quantum of electromagnetic radiation. The energy $E$ of a photon is given by the relation $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency of the radiation.
Since the frequency $\nu = \frac{c}{\lambda}$,where $c$ is the speed of light and $\lambda$ is the wavelength,we can substitute this into the energy equation.
Therefore,$E = \frac{hc}{\lambda}$.
This is the correct expression for the energy of a photon.

(C) The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Also,the relationship between the speed of light $c$,frequency $\nu$,and wavelength $\lambda$ is given by $c = \nu \lambda$.
Rearranging this formula to solve for wavelength $\lambda$,we get $\lambda = \frac{c}{\nu}$.

(D) The energy of a photon is given by the formula $E = h\nu$.
Given frequency $\nu = 10^{12} \text{ MHz} = 10^{12} \times 10^6 \text{ Hz} = 10^{18} \text{ Hz}$.
Planck's constant $h \approx 4.14 \times 10^{-15} \text{ eV} \cdot \text{s}$.
Therefore,$E = (4.14 \times 10^{-15} \text{ eV} \cdot \text{s}) \times (10^{18} \text{ Hz}) = 4.14 \times 10^3 \text{ eV}$.
Thus,the correct option is $D$.

(C) The energy $E$ of a beam of light containing $n$ photons of frequency $\nu$ is given by $E = n h \nu$,where $h$ is Planck's constant.
Since both beams are equally energetic,we have $E_1 = E_2$.
Therefore,$n_1 h \nu_1 = n_2 h \nu_2$.
Dividing both sides by $h$,we get $n_1 \nu_1 = n_2 \nu_2$.
Rearranging the terms to find the ratio $\frac{n_1}{n_2}$,we get $\frac{n_1}{n_2} = \frac{\nu_2}{\nu_1}$.

(B) The energy $E$ of a photon is given by the formula $E = h\nu$,where $h$ is Planck's constant $(h \approx 6.62 \times 10^{-34} \ J \cdot s)$ and $\nu$ is the frequency of the radiation.
Given: $\nu = 1.0 \times 10^{14} \ Hz$.
Substituting the values: $E = (6.62 \times 10^{-34} \ J \cdot s) \times (1.0 \times 10^{14} \ Hz)$.
$E = 6.62 \times 10^{-34 + 14} \ J$.
$E = 6.62 \times 10^{-20} \ J$.

(A) The power $P$ of the transmitter is the energy emitted per second. The energy of a single photon is given by $E = h\nu$,where $h = 6.6 \times 10^{-34} \, J \cdot s$ is Planck's constant and $\nu$ is the frequency.
If $n$ is the number of photons emitted per second,then the total power is $P = n \times E = n \times h\nu$.
Therefore,$n = \frac{P}{h\nu}$.
Given: $P = 10 \, kW = 10^4 \, W$ and $\nu = 880 \, kHz = 880 \times 10^3 \, Hz$.
Substituting the values:
$n = \frac{10^4}{(6.6 \times 10^{-34}) \times (880 \times 10^3)}$
$n = \frac{10^4}{5808 \times 10^{-31}}$
$n = \frac{10^4}{5.808 \times 10^{-28}}$
$n \approx 1.72 \times 10^{31}$ photons per second.

(C) The power $P$ radiated by the lamp is given by the energy of $n$ photons emitted in time $t$: $P = \frac{n}{t} \frac{hc}{\lambda}$.
Rearranging for the number of photons per second $\frac{n}{t} = \frac{P \lambda}{hc}$.
Given: $P = 100 \ W$,$\lambda = 5000 \ \mathring{A} = 5000 \times 10^{-10} \ m$,$h = 6.63 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$.
Substituting the values: $\frac{n}{t} = \frac{100 \times 5000 \times 10^{-10}}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
$\frac{n}{t} = \frac{5 \times 10^{-5}}{19.89 \times 10^{-26}} \approx 2.51 \times 10^{20}$.
Thus,the number of photons radiated per second is approximately $2.5 \times 10^{20}$.

(C) When a photon collides with an electron (Compton scattering),it transfers some of its energy and momentum to the electron.
According to the conservation of energy and momentum,the energy $(E = h\nu)$ and momentum $(p = h/\lambda)$ of the photon decrease after the collision.
However,the speed of a photon in a vacuum is a universal constant,$c = 3 \times 10^8 \ m/s$,which remains unchanged regardless of the interaction.
Therefore,only the speed of the photon remains constant.

(C) The power $P$ radiated by the transmitter is given by $P = \frac{E}{t}$, where $E$ is the total energy emitted in time $t$.
Since $E = n \times E_{\text{photon}} = n \times \frac{hc}{\lambda}$, where $n$ is the number of photons emitted per second, we have $P = \frac{nhc}{\lambda}$.
Given: $P = 1 \text{ kW} = 10^3 \text{ W}$, $\lambda = 198.6 \text{ m}$, $h = 6.63 \times 10^{-34} \text{ J s}$, and $c = 3 \times 10^8 \text{ m/s}$.
Substituting the values: $10^3 = \frac{n \times 6.63 \times 10^{-34} \times 3 \times 10^8}{198.6}$.
$n = \frac{10^3 \times 198.6}{6.63 \times 10^{-34} \times 3 \times 10^8} \approx \frac{198600}{19.89} \times 10^{26} \approx 10^{30}$.
Thus, the number of photons emitted per second is $10^{30}$.

(C) The power $P$ of the bulb is the total energy emitted per second,given by $P = \frac{nE}{\Delta t}$,where $n$ is the number of photons emitted per second and $E = \frac{hc}{\lambda}$ is the energy of one photon.
Given: $P = 100 \ W$,$\lambda = 540 \ nm = 540 \times 10^{-9} \ m$,$h = 6 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$.
Substituting these values into the formula $P = \frac{nhc}{\lambda}$:
$100 = \frac{n \times (6 \times 10^{-34}) \times (3 \times 10^8)}{540 \times 10^{-9}}$
$100 = \frac{n \times 18 \times 10^{-26}}{540 \times 10^{-9}}$
$100 = n \times \frac{18}{540} \times 10^{-17}$
$100 = n \times \frac{1}{30} \times 10^{-17}$
$n = 100 \times 30 \times 10^{17} = 3000 \times 10^{17} = 3 \times 10^{20}$
Thus,the number of photons emitted per second is $3 \times 10^{20}$.

(C) The energy of a photon is given by the relation $E = \frac{hc}{\lambda}$.
Since $h$ and $c$ are constants,we have $E \propto \frac{1}{\lambda}$.
Therefore,$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Given $\lambda_1 = 6000 \ \mathring A$,$E_1 = 3.32 \times 10^{-19} \ J$,and $\lambda_2 = 4000 \ \mathring A$.
Substituting the values: $\frac{3.32 \times 10^{-19}}{E_2} = \frac{4000}{6000} = \frac{2}{3}$.
$E_2 = \frac{3}{2} \times 3.32 \times 10^{-19} \ J = 4.98 \times 10^{-19} \ J$.
To convert energy from Joules to $eV$,we divide by $1.6 \times 10^{-19} \ J/eV$:
$E_2 = \frac{4.98 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV \approx 3.11 \ eV$.
Thus,the correct option is $C$.

(D) The velocity of a photon in a vacuum is a constant value,denoted by $c$,which is approximately $3 \times 10^8 \ m/s$.
Since the velocity of light $(c)$ is independent of the frequency $(\nu)$ and wavelength $(\lambda)$ of the photon,it can be expressed as being proportional to $\nu^0$ (i.e.,$c \propto \nu^0 = 1$).
Therefore,the velocity does not change with frequency.

(C) The energy of a photon is given by $E = h\nu = \frac{hc}{\lambda}$.
The momentum of a photon is given by $p = \frac{h}{\lambda}$.
By dividing these two equations,we get $E = pc$,where $c$ is the speed of light.
Since $c$ is a constant,the momentum $p$ is directly proportional to the energy $E$ $(p \propto E)$.
If the energy $E$ is increased by a factor of $4$,the momentum $p$ will also increase by a factor of $4$.

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

(D) The photoelectric effect cannot be explained by the classical wave theory of light because the wave theory predicts that the energy of emitted electrons should depend on the intensity of light,whereas experiments show it depends on the frequency. Albert Einstein explained the photoelectric effect by proposing that light consists of discrete packets of energy called photons,where the energy of each photon is given by $E = h\nu$. This quantum nature of light successfully explains why the kinetic energy of emitted electrons depends on the frequency of incident light.

(A) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Using the shortcut formula $E(\text{eV}) = \frac{12375}{\lambda(\mathring A)}$,where $\lambda = 5000 \mathring A$:
$E = \frac{12375}{5000} \text{ eV}$.
$E = 2.475 \text{ eV}$.
Rounding to the nearest option,we get $E \approx 2.5 \text{ eV}$.

(A) For a photon,the energy $E$ is related to its momentum $p$ by the equation $E = pc$,where $c$ is the speed of light in vacuum.
Squaring both sides of this equation,we get $E^2 = (pc)^2$,which simplifies to $E^2 = p^2c^2$.
Therefore,the correct relation is $E^2 = p^2c^2$.

(B) For electron-positron pair production,the minimum energy required is $1.02 \ MeV$.
The energy of the given photon is $E = 1.7 \times 10^{-13} \ J$.
To convert this energy into $MeV$,we divide by the charge of an electron $(1.6 \times 10^{-19} \ C)$:
$E = \frac{1.7 \times 10^{-13}}{1.6 \times 10^{-19}} \ eV = 1.0625 \times 10^6 \ eV = 1.0625 \ MeV$.
Since the energy of the photon $(1.0625 \ MeV)$ is greater than the threshold energy required for pair production $(1.02 \ MeV)$,an electron-positron pair will be created.

(B) Given: Rest mass energy $E_0 = m_0c^2 = 0.51 \ MeV$ and velocity $v = 0.8 \ c$.
The kinetic energy $(K.E.)$ of a relativistic particle is given by $K.E. = mc^2 - m_0c^2$.
The relativistic mass $m$ is given by $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$.
Substituting $v = 0.8 \ c$:
$m = \frac{m_0}{\sqrt{1 - (0.8)^2}} = \frac{m_0}{\sqrt{1 - 0.64}} = \frac{m_0}{\sqrt{0.36}} = \frac{m_0}{0.6}$.
Now,the total energy $E = mc^2 = \frac{m_0c^2}{0.6} = \frac{0.51 \ MeV}{0.6} = 0.85 \ MeV$.
Therefore,$K.E. = E - E_0 = 0.85 \ MeV - 0.51 \ MeV = 0.34 \ MeV$.

(C) The intensity $I$ is defined as power per unit area: $I = \frac{P}{A}$.
Given $I = 10^{-10} \ W/m^2$ and $A = 10^{-6} \ m^2$,the power $P$ entering the eye is $P = I \times A = 10^{-10} \times 10^{-6} = 10^{-16} \ W$.
The power of a beam of $n$ photons per second is given by $P = \frac{n \cdot h \cdot c}{\lambda}$.
Rearranging for the number of photons per second $\frac{n}{t} = \frac{P \cdot \lambda}{h \cdot c}$.
Substituting the values: $\frac{n}{t} = \frac{10^{-16} \times 5.6 \times 10^{-7}}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
$\frac{n}{t} = \frac{5.6 \times 10^{-23}}{19.89 \times 10^{-26}} \approx 281.5 \approx 300$ photons per second.

(A) The energy of a single photon of green light is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 12400 \ \text{eV} \ \mathring{A}$, we get $E = \frac{12400}{5000} \ \text{eV} = 2.48 \ \text{eV}$.
Converting this to Joules: $E = 2.48 \times 1.6 \times 10^{-19} \ J \approx 3.97 \times 10^{-19} \ J \approx 4 \times 10^{-19} \ J$.
The minimum intensity detected by the eye $(I_{Eye})$ is the product of photon flux and energy per photon:
$I_{Eye} = (5 \times 10^4 \ \text{photons}/m^2s) \times (4 \times 10^{-19} \ J) = 2 \times 10^{-14} \ W/m^2$.
The sensitivity of a detector is inversely proportional to the minimum intensity it can detect.
Therefore, the ratio of sensitivity is $\frac{S_{Eye}}{S_{Ear}} = \frac{I_{Ear}}{I_{Eye}} = \frac{10^{-13}}{2 \times 10^{-14}} = \frac{10}{2} = 5$.
Thus, the eye is $5$ times more sensitive than the ear.

(A) In a vacuum,the velocity of a photon (i.e.,light) is a constant value,$c \approx 3 \times 10^8 \ m/s$,regardless of its frequency or wavelength. Since the velocity does not depend on the frequency,the graph of velocity versus frequency is a straight line parallel to the frequency axis. The graph is shown below:
Velocity of photon $(c)$ vs Frequency $(\nu)$: $A$ horizontal line at $y = c$.

(B) $Photon$ is an elementary particle,which is a quantum of electromagnetic radiation. It is considered a fundamental particle in the Standard Model of particle physics and is non-divisible. In contrast,an $Atom$ consists of a nucleus and electrons,a $Nucleus$ consists of protons and neutrons,and a $Proton$ is a composite particle made of quarks. Therefore,the correct option is $B$.