Particle Nature of Light : Photon Questions in English

(D) The energy 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.
According to the de-Broglie relation,the momentum $p$ is given by $p = \frac{h}{\lambda}$,which implies $\lambda = \frac{h}{p}$.
Substituting $\lambda$ into the energy equation: $E = \frac{hc}{h/p} = pc$.
Therefore,the momentum is $p = \frac{E}{c}$.
Given: $E = 1 \text{ MeV} = 1 \times 10^{6} \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-13} \text{ J}$ and $c = 3 \times 10^{8} \text{ m/s}$.
Substituting these values: $p = \frac{1.6 \times 10^{-13}}{3 \times 10^{8}} \approx 0.533 \times 10^{-21} \text{ kg-m/s} = 5.33 \times 10^{-22} \text{ kg-m/s}$.
Rounding to the nearest provided option,we get $p = 5 \times 10^{-22} \text{ kg-m/s}$.

(D) The power $P$ of the bulb is the total energy emitted per second. 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 $P = n \times E = n \frac{hc}{\lambda}$.
Rearranging for $n$,we get $n = \frac{P \lambda}{hc}$.
Given: $P = 66 \ W$,$\lambda = 600 \ nm = 600 \times 10^{-9} \ m$,$h = 6.6 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$.
Substituting the values:
$n = \frac{66 \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}$
$n = \frac{66 \times 600 \times 10^{-9}}{19.8 \times 10^{-26}}$
$n = \frac{39600 \times 10^{-9}}{19.8 \times 10^{-26}}$
$n = 2000 \times 10^{17} = 2 \times 10^{20}$.
Thus,the number of photons emitted per second is $2 \times 10^{20}$.

(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Since $h$ and $c$ are constants,the energy $E$ is inversely proportional to the wavelength $\lambda$,i.e.,$E \propto \frac{1}{\lambda}$.
Therefore,we can write the ratio as $\frac{E_{2}}{E_{1}} = \frac{\lambda_{1}}{\lambda_{2}}$.
Given: $E_{1} = 3.2 \times 10^{-19} \ J$,$\lambda_{1} = 6000 \mathring{A}$,and $\lambda_{2} = 4000 \mathring{A}$.
Substituting the values: $\frac{E_{2}}{3.2 \times 10^{-19}} = \frac{6000}{4000} = \frac{6}{4} = 1.5$.
$E_{2} = 1.5 \times 3.2 \times 10^{-19} \ J$.
$E_{2} = 4.80 \times 10^{-19} \ J$.

(A) Given:
Energy $E = 35 \text{ keV} = 35 \times 10^{3} \times 1.6 \times 10^{-19} \text{ J} = 56 \times 10^{-16} \text{ J}$.
Using the formula for the wavelength of a photon:
$E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$
Substituting the values:
$\lambda = \frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{35 \times 10^{3} \times 1.6 \times 10^{-19}}$
$\lambda = \frac{19.875 \times 10^{-26}}{56 \times 10^{-16}}$
$\lambda \approx 0.3549 \times 10^{-10} \text{ m} \approx 0.355 \times 10^{-10} \text{ m} = 35.5 \times 10^{-12} \text{ m}$.
Rounding to the nearest provided option,the correct value is $35 \times 10^{-12} \text{ m}$.

(B) The energy of a photon is given by $E = hf$,where $h$ is Planck's constant and $f$ is the frequency.
According to Einstein's mass-energy equivalence relation,$E = mc^2$.
Equating the two expressions for energy: $mc^2 = hf$.
Since momentum $p = mc$,we can write $mc^2 = (mc)c = pc = hf$.
Therefore,the momentum $p = \frac{hf}{c}$.

(D) The energy of a photon is given by the formula $E = h \nu$,where $h$ is Planck's constant $(6.63 \times 10^{-34} \,J \cdot s)$ and $\nu$ is the frequency of the light.
Given: $\nu = 6 \times 10^{14} \,Hz$.
Substituting the values: $E = (6.63 \times 10^{-34} \,J \cdot s) \times (6 \times 10^{14} \,Hz) = 3.978 \times 10^{-19} \,J$.
To convert the energy from Joules to electron-volts $(eV)$,we divide by the charge of an electron $(1.6 \times 10^{-19} \,C)$:
$E = \frac{3.978 \times 10^{-19} \,J}{1.6 \times 10^{-19} \,J/eV} \approx 2.486 \,eV$.
Rounding to the nearest value,we get $E \approx 2.5 \,eV$.

(D) For a proton,the total energy $E_{total} = E_k + mc^2 = \frac{1}{8}mc^2 + mc^2 = \frac{9}{8}mc^2$.
Using the relativistic relation $E_{total}^2 = (pc)^2 + (mc^2)^2$:
$(\frac{9}{8}mc^2)^2 = (p_1c)^2 + (mc^2)^2$
$\frac{81}{64}m^2c^4 = p_1^2c^2 + m^2c^4$
$p_1^2c^2 = (\frac{81}{64} - 1)m^2c^4 = \frac{17}{64}m^2c^4$
$p_1 = \frac{\sqrt{17}}{8}mc$.
For a photon,energy $E = p_2c = E_k = \frac{1}{8}mc^2$,so $p_2 = \frac{mc}{8}$.
The ratio $\frac{p_1 - p_2}{p_1} = 1 - \frac{p_2}{p_1} = 1 - \frac{mc/8}{\sqrt{17}mc/8} = 1 - \frac{1}{\sqrt{17}}$.
Note: If we assume non-relativistic kinetic energy $E_k = \frac{p_1^2}{2m} = \frac{1}{8}mc^2$,then $p_1^2 = \frac{1}{4}m^2c^2$,so $p_1 = \frac{mc}{2}$.
Then $p_2 = \frac{E_k}{c} = \frac{mc}{8}$.
The ratio $\frac{p_1 - p_2}{p_1} = \frac{mc/2 - mc/8}{mc/2} = \frac{3/8}{1/2} = \frac{3}{4}$.

(B) Given: Power of the source $P = 60 \,W$.
Wavelength of light $\lambda = 662.5 \,nm = 6.625 \times 10^{-7} \,m$.
The energy of a single photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values ($h = 6.625 \times 10^{-34} \,J \cdot s$ and $c = 3 \times 10^8 \,m/s$):
$E = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6.625 \times 10^{-7}} = 3 \times 10^{-19} \,J$.
The number of photons emitted per second $(n)$ is given by the ratio of total power to the energy of one photon:
$n = \frac{P}{E} = \frac{60}{3 \times 10^{-19}} = 20 \times 10^{19} = 2 \times 10^{20}$ photons/second.

(A) The momentum of a single photon with wavelength $\lambda$ is given by $p = \frac{h}{\lambda}$.
Since $n$ photons are absorbed by the black body,the total momentum transferred to the body is the sum of the momenta of all $n$ photons.
Therefore,the total momentum gained by the body is $P = n \times p = n \times \frac{h}{\lambda} = \frac{nh}{\lambda}$.

(D) $LASER$ action relies on the following fundamental processes:
$(i)$ Population inversion: Achieving a state where more atoms are in an excited state than in the ground state.
(ii) Stimulated emission: An incident photon triggers an excited atom to emit a second photon identical in phase,frequency,and direction.
(iii) Amplification: The process of increasing the intensity of light by repeating stimulated emission within an optical cavity.
Therefore,all the given options are correct.

(A) According to the second postulate of Einstein's Special Theory of Relativity,the speed of light in a vacuum is a universal constant $c$ $(3 \times 10^8 \ m/s)$.
This speed is independent of the motion of the source or the motion of the observer.
Therefore,regardless of the velocity $v$ at which the source moves away from the observer,the relative velocity of light with respect to the observer remains $c$.

(B) The power $P$ of the laser 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 one photon is $E = h\nu$,where $h = 6.6 \times 10^{-34} \,J \,s$ and $\nu = 5 \times 10^{14} \,Hz$.
$E = (6.6 \times 10^{-34}) \times (5 \times 10^{14}) = 33 \times 10^{-20} \,J$.
The power $P = 33 \,mW = 33 \times 10^{-3} \,W$.
Number of photons per second $n = P / E$.
$n = (33 \times 10^{-3}) / (33 \times 10^{-20}) = 10^{17} = 10 \times 10^{16}$ photons per second.
Thus,the correct option is $B$.

(A) Given: Wavelength $\lambda = 4500 \ \text{Å} = 4500 \times 10^{-10} \ \text{m}$,Power $P = 150 \ \text{W}$,Efficiency $\eta = 8 \% = 0.08$.
The power of the light emitted by the lamp is $P_{\text{out}} = P \times \eta = 150 \times 0.08 = 12 \ \text{W}$.
The energy of a single photon is $E = \frac{hc}{\lambda}$.
The number of photons emitted per second $n$ is given by $n = \frac{P_{\text{out}}}{E} = \frac{P_{\text{out}} \times \lambda}{hc}$.
Using $hc \approx 1240 \ \text{eV} \cdot \text{nm} = 12400 \ \text{eV} \cdot \text{Å}$ and $1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J}$,we have $hc = 12400 \times 1.6 \times 10^{-19} \ \text{J} \cdot \text{Å} = 1.984 \times 10^{-15} \ \text{J} \cdot \text{m}$.
Substituting the values: $n = \frac{12 \times 4500 \times 10^{-10}}{6.626 \times 10^{-34} \times 3 \times 10^8} \approx 2.717 \times 10^{19} \text{ (or using } hc \approx 1.987 \times 10^{-16} \text{ J} \cdot \text{m, } n \approx 2.717 \times 10^{19} \text{)}$.
Re-evaluating with standard constants: $n = \frac{12 \times 4500 \times 10^{-10}}{6.63 \times 10^{-34} \times 3 \times 10^8} = 2.715 \times 10^{19}$. Given the options,the calculation $27.17 \times 10^{18}$ is the intended answer.

(B) The power $P$ of the transmitter is given by $P = \frac{n E_{photon}}{t}$,where $n$ is the number of photons emitted in time $t$ and $E_{photon} = \frac{hc}{\lambda}$.
Thus,the number of photons emitted per second is $\frac{n}{t} = \frac{P \lambda}{hc}$.
Given: $P = 10 \text{ kW} = 10^4 \text{ W}$,$\lambda = 500 \text{ m}$,$h = 6.63 \times 10^{-34} \text{ J s}$,and $c = 3 \times 10^8 \text{ m/s}$.
Substituting the values:
$\frac{n}{t} = \frac{10^4 \times 500}{6.63 \times 10^{-34} \times 3 \times 10^8}$
$\frac{n}{t} = \frac{5 \times 10^6}{19.89 \times 10^{-26}}$
$\frac{n}{t} \approx 0.251 \times 10^{32} = 2.51 \times 10^{31}$.
Therefore,the order of magnitude is $10^{31}$.

(C) photon is a quantum of electromagnetic radiation. According to the theory of relativity,the rest mass of a photon is $0$. Photons travel at the speed of light $c$ in a vacuum and possess energy $E = h\nu$ and momentum $p = h/\lambda$,but they do not have any rest mass.

(A) The power $P$ of the laser is the total energy emitted per second. The energy of a single photon is given by $E = \frac{hc}{\lambda}$.
Therefore,the number of photons emitted per second $n$ is given by $n = \frac{P}{E} = \frac{P \lambda}{hc}$.
Given values: $P = 6.6 \times 10^{-3} \,W$,$\lambda = 600 \times 10^{-9} \,m$,$h = 6.6 \times 10^{-34} \,J \cdot s$,and $c = 3 \times 10^8 \,m/s$.
Substituting these values into the formula:
$n = \frac{(6.6 \times 10^{-3} \,W) \times (600 \times 10^{-9} \,m)}{(6.6 \times 10^{-34} \,J \cdot s) \times (3 \times 10^8 \,m/s)}$
$n = \frac{6.6 \times 600 \times 10^{-12}}{6.6 \times 3 \times 10^{-26}}$
$n = \frac{600 \times 10^{-12}}{3 \times 10^{-26}}$
$n = 200 \times 10^{14} = 2 \times 10^{16}$ photons per second.

(A) The wavelength of the photon is given as $\lambda = 4000 \text{ Å} = 4000 \times 10^{-10} \text{ m} = 4 \times 10^{-7} \text{ m}$.
The energy $E$ of a photon is calculated using the formula $E = \frac{hc}{\lambda}$,where $h = 6.626 \times 10^{-34} \text{ J s}$ is Planck's constant and $c = 3 \times 10^8 \text{ m/s}$ is the speed of light.
Substituting the values:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}}$
$E = \frac{19.878 \times 10^{-26}}{4 \times 10^{-7}}$
$E = 4.9695 \times 10^{-19} \text{ J}$
Rounding to the nearest provided option,we get $E \approx 4.95 \times 10^{-19} \text{ J}$.

(B) positron,also known as an antielectron,is the antiparticle of the electron. It has the same mass as an electron but carries an opposite electric charge ($+e$ instead of $-e$).

(C) photon is a quantum of electromagnetic radiation. According to the quantum theory of light, the energy of a photon is given by $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of the radiation.
Since the energy depends only on the frequency, statement $C$ is incorrect because it claims energy depends on intensity.
Photons are massless particles that travel at the speed of light $(c \approx 3 \times 10^8 \ m/s)$.
Because photons are electrically neutral, they do not experience any force in electric or magnetic fields, meaning they are not deflected by them.

(D) The intensity $I$ at a distance $r = 5 \ m$ from the source is given by $I = \frac{P}{4 \pi r^2}$.
Substituting the values: $I = \frac{942}{4 \times 3.14 \times 5^2} = \frac{942}{12.56 \times 25} = \frac{942}{314} = 3 \ W/m^2$.
The energy of a single photon is $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} = \frac{19.8 \times 10^{-26}}{6.6 \times 10^{-7}} = 3 \times 10^{-19} \ J$.
The photon flux $\phi$ is the number of photons per unit area per unit time,given by $\phi = \frac{I}{E}$.
$\phi = \frac{3}{3 \times 10^{-19}} = 1 \times 10^{19} \ \frac{\text{photon}}{m^2 \cdot s}$.

(C) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} = 3 \times 10^{-19} J$.
The total energy emitted by the lamp in time $t = 100 ms = 0.1 s$ is $E_{total} = P \times t = 300 \times 0.1 = 30 J$.
The intensity at a distance $r = 10 m$ is distributed over a sphere of area $A_{sphere} = 4\pi r^2 = 4\pi(10)^2 = 400\pi m^2$.
The area of the sensor opening with radius $r_s = 1 cm = 10^{-2} m$ is $A_{sensor} = \pi r_s^2 = \pi(10^{-2})^2 = \pi \times 10^{-4} m^2$.
The energy incident on the sensor is $E_{sensor} = E_{total} \times \frac{A_{sensor}}{A_{sphere}} = 30 \times \frac{\pi \times 10^{-4}}{400\pi} = \frac{30 \times 10^{-4}}{400} = 0.075 \times 10^{-4} = 7.5 \times 10^{-6} J$.
The number of photons $n$ is $\frac{E_{sensor}}{E} = \frac{7.5 \times 10^{-6}}{3 \times 10^{-19}} = 2.5 \times 10^{13}$.

(D) For the Compton effect,the shift in wavelength is given by the relation:
$\Delta \lambda = \lambda_2 - \lambda_1 = \frac{h}{m_0 c}(1 - \cos \theta)$
Here,$\lambda_1$ is the wavelength of incident radiation,$\lambda_2$ is the wavelength of scattered radiation,and $\frac{h}{m_0 c} \approx 0.024 \ \text{Å}$ is the Compton wavelength.
Given: $\lambda_2 = 0.22 \ \text{Å}$ and $\theta = 60^{\circ}$.
Substituting the values:
$0.22 - \lambda_1 = 0.024(1 - \cos 60^{\circ})$
$0.22 - \lambda_1 = 0.024(1 - 0.5)$
$0.22 - \lambda_1 = 0.024 \times 0.5$
$0.22 - \lambda_1 = 0.012$
$\lambda_1 = 0.22 - 0.012 = 0.208 \ \text{Å}$

(B) Bose-Einstein statistics describes the statistical behavior of particles known as bosons.
Bosons are particles that have integral spin,i.e.,spin $s = 0, 1, 2, \dots$.
These particles do not obey the Pauli exclusion principle,meaning multiple particles can occupy the same quantum state.
In contrast,Fermi-Dirac statistics applies to fermions,which have half-odd integral spin $(s = 1/2, 3/2, 5/2, \dots)$ and obey the Pauli exclusion principle.
Therefore,Bose-Einstein statistics is applicable to particles with integral spin.

(A) Intensity of radiation is defined as the energy incident per unit area per unit time: $I = \frac{E}{A \cdot t}$.
For a photon of frequency $f$,the energy is $E = hf$. Thus,the total energy incident is $E_{total} = N \cdot hf$.
For surface $A$: $I_A = \frac{n \cdot hf}{A \cdot t}$.
For surface $B$: $I_B = \frac{(2n) \cdot h(3f)}{A \cdot (4t)} = \frac{6nhf}{4At} = \frac{3nhf}{2At}$.
Taking the ratio: $\frac{I_A}{I_B} = \frac{nhf}{At} \cdot \frac{2At}{3nhf} = \frac{2}{3}$.
Therefore,the ratio is $2: 3$.

(A) The power $P$ of the lamp is given by the total energy emitted per second,which is the product of the number of photons emitted per second $N$ and the energy of a single photon $E = \frac{hc}{\lambda}$.
Given: $N = 10^{20} \ s^{-1}$,$\lambda = 660 \ nm = 660 \times 10^{-9} \ m$,$h = 6.6 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$.
$P = N \times \frac{hc}{\lambda}$
$P = 10^{20} \times \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}}$
$P = 10^{20} \times \frac{19.8 \times 10^{-26}}{660 \times 10^{-9}}$
$P = 10^{20} \times 0.03 \times 10^{-17}$
$P = 30 \ W$.

(D) The power $P$ of the laser source is given by the formula $P = n \times E$,where $n$ is the number of photons emitted per second and $E$ is the energy of a single photon.
Energy of one photon is $E = \frac{hc}{\lambda}$.
Substituting this into the power equation: $P = \frac{n \times hc}{\lambda}$.
Given values: $P = 6 \ mW = 6 \times 10^{-3} \ W$,$\lambda = 663 \ nm = 663 \times 10^{-9} \ m$,$h = 6.63 \times 10^{-34} \ J.s$,and $c = 3 \times 10^{8} \ m/s$.
Rearranging for $n$: $n = \frac{P \times \lambda}{h \times c}$.
$n = \frac{6 \times 10^{-3} \times 663 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^{8}}$.
$n = \frac{6 \times 663 \times 10^{-12}}{19.89 \times 10^{-26}}$.
$n = \frac{3978 \times 10^{-12}}{19.89 \times 10^{-26}} = 200 \times 10^{14} = 2 \times 10^{16}$ photons per second.

(A) The power $P$ emitted by a source is given by the product of the number of photons emitted per second $(n)$ and the energy of a single photon $(E = h\nu)$.
Given: Power $P = 4 \times 10^{-3} \text{ W}$,frequency $\nu = 6 \times 10^{14} \text{ Hz}$,and Planck's constant $h = 6.63 \times 10^{-34} \text{ Js}$.
The formula is $P = n \cdot h\nu$.
Rearranging for $n$: $n = \frac{P}{h\nu}$.
Substituting the values: $n = \frac{4 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}}$.
$n = \frac{4 \times 10^{-3}}{39.78 \times 10^{-20}} = \frac{4}{39.78} \times 10^{17} \approx 0.10055 \times 10^{17} = 1.0055 \times 10^{16}$.
Rounding to the nearest significant figure,we get $n \approx 1 \times 10^{16} \text{ photons per second}$.