Modern Physics

The dual nature of matter and radiation refers to the concept that particles of matter exhibit both particle-like and wave-like properties. This duality is a fundamental aspect of modern Physics. Electromagnetic radiation, such as light, behaves like a wave with characteristics such as wavelength, frequency, and interference patterns. However, it can also be described as discrete packets of energy called photons, each carrying energy proportional to its frequency. In the early 20th century by pioneers like Max Planck, Albert Einstein, and Erwin Schrödinger, quantum mechanics describes particles and radiation with wave functions that evolve deterministically and probabilistically.

1.0Dual Nature Of Light

Light does not have a definite nature; rather, its nature depends on its experimental phenomenon. This is known as the dual nature of light. 

2.0Types of Emission

Emission- It is a process of emission of electrons from a metal surface.

1. Thermionic Emission
2. Field Emission
3. Photoelectric Emission

Work Function($\varphi$) - It is the minimum energy an electron requires to escape from a metal surface. It is measured in electron volts(eV).

3.0Einstein’s Quantum Theory of Light

Light behaves as Quanta. These energy quanta are called Photons, and they have definite Energy and Momentum and travel with the speed of light in a vacuum.

4.0Energy of Photon

• Energy is radiated from the source in the form of small packets called photons. 
• According to Planck's Theory: Energy(E)  $\propto$ (Frequency)
• $E=hv=\frac{hc}{\lambda }\frac{12400}{\lambda }eV-\mathring{A}$

5.0Linear Momentum of Photon

$\mathrm{p}=\frac{E}{c}=\frac{hv}{c}=\frac{h}{\lambda }$

6.0Effective Mass of Photon

• $\mathrm{m}=\frac{E}{c^2}=\frac{hc}{c^2\lambda }=\frac{hc}{\lambda }$
• The rest mass of the Photon is always zero.
• ${\lambda }_R>{\lambda }_V$

7.0Intensity of Light

• I=EAt=PA
• S.I Unit:Joulem2-s or wattm2
• I=N(h)At=n(h)A    (∴n=Nt, no.of photons per sec.)

8.0Radiation Force and Pressure

• Radiation Force: $\mathrm{F}=\mathrm{n}\frac{2h}{\lambda }\text{ here }\mathrm{n}=\frac{p\lambda }{hc}$
• Radiation Pressure: $\mathrm{P}=\frac{F}{A}=\frac{2P}{cA}=\frac{2I}{c}\left[\because |I=\frac{P}{A}\right]$

9.0Photoelectric Effect

It is a phenomenon of ejecting electrons by falling light of a suitable frequency or wavelength on a metal surface. Ejected electrons are called photoelectrons, and the current flowing due to these photoelectrons is called photoelectric current.

10.0Saturation current

When all the photoelectrons emitted by the cathode reach the anode then current flowing in the circuit at that instant is known as saturation current; this is the maximum value of Photoelectric current.

11.0Stopping Potential

The minimum magnitude of the negative potential of the anode concerning the cathode for which the current is zero is called the stopping potential or cut-off voltage; this voltage is independent of intensity.

12.0Graph of Various Quantities

Graph of Various Quantities Observed from Lenard’s Experiment of Photoelectric Effect:

1. Photoelectric current versus Intensity of Light

2. Photoelectric Current versus stopping Potential (Frequency=Constant)

3. Photoelectric Current versus Stopping Potential (Frequency = variable)

4. Stopping Potential Versus Frequency

13.0Einstein Photoelectric Equation               

• The surface absorbs radiation in the form of photons. The Energy of each Photon depends on its frequency. At a time, only one Photon can interact with one electron.
• The Energy of the photon used by the electron is,

h$v$ = K.E of electron + Energy required to make electron free from the metal surface(0) + Energy lost in collision before emission (Q)

∴ If Q = 0, means there is no heat loss, then K.E of electron is maximum.

$\mathrm{hv}=K.E_{\text{Max }}+{\varphi }_0\dots$….Einstein Photoelectric Equation

$K\cdot E_{\text{Max }}=\mathrm{hu}-{\varphi }_0$

$e{V}_0=hv-h{v}_0$ ($\therefore v_0$ is the threshold frequency for that $V_0=0$)

• K.E. and Stopping Potential $\left(V_0\right)$ depend on frequency.
• Kinetic Energy cannot be negative so that $hu>h{v}_0$
•  $v>v_0$ (It means if frequency is less than ${}^{\prime }{v}_0^{\prime }$, electrons do not come out.

The Graph between K.E. and Frequency

14.0Quantum Efficiency

$\text{ Quantum Efficiency }(\mathrm{x})=\frac{\text{ number of electrons emitted per second }}{\text{ total number of photons incident per second }}=\frac{n_e}{n_{\text{ph }}}$

• If quantum efficiency is x % then $n_e=\frac{x}{100}\left[\because n_{ph}=\left(5\times 10^{24}{\mathrm{J}}^{-1}{\mathrm{m}}^{-1}\right)\mathrm{P}\lambda \right]$

15.0Matter Waves and De Broglie Hypothesis

• Waves associated with a moving particle are matter waves.
• Matter waves are probabilistic because they can be expressed as probability density functions.

The Formula for de-Broglie wavelength

$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}$

De-Broglie wavelength for a charged particle

$\lambda =\frac{h}{\sqrt{2mqV}}$

16.0Explanation of Bohr Quantisation Condition

According to De Broglie, electrons revolve around the nucleus in the form of stationary waves; electrons revolve in those circular orbits whose circumference is an integral multiple of the de-Broglie wavelength associated with the electrons.

$2\pi r=n\lambda$

$2\pi r=n\frac{h}{mv}$

$\mathrm{mvr}=\frac{nh}{2\pi }$…… This is Bohr quantisation Condition

$r_n=(0.529\mathring{A})\frac{n^2}{Z}(\therefore Z=\text{ Atomic number })$

17.0Nucleus and Types of Nuclei

• The Nucleus is a central core of every atom discovered by Rutherford in scattering experiment.
• Order of nuclear size=10^-15m and of atomic size=10^-10 m
• Atomic mass is expressed by atomic mass unit (u)
• 1u = 1.66 x 10^-27 kg

18.0Types of Nuclei

• Isotope- same Z
• Isobar-same A
• Isotone-same(A-Z)
• Size of Nucleus : $\mathrm{R}={\mathrm{R}}_0{A}^{1/3}$ here,  $R_0=EmpiricalConstant=1.2fermi$
• Volume $\propto$ Mass Number (A)
• $\text{ Density of Nucleus }=\frac{3m_p}{4\pi R_0^3}=2.3\times 10^{17}\mathrm{kg}/{\mathrm{m}}^3$

19.0Nuclear Forces and Its Properties

• The most robust interaction holding nucleons together to form nuclei is strong enough to overcome the electric repulsion of protons and neutrons.
• The most vital force in the universe.
• Works only between the nucleons
• Very short range force
• It depends on distance
• Charge independent
• Spin dependent 
• Non central force
• Nature-(1) Attractive-distance greater than 0.8 fm (2) Repulsive-distance lesser than 0.8 fm.

20.0Mass Defect and Nuclear Binding Energy

• Mass of a nucleus is always less than the sum of masses of its constituent nucleons;this difference is called mass defect

$\Delta$m = [Zm_p + (A - Z)M_n] - M

• Binding energy (E_b) is the energy required to split it into its nucleons: ${\mathrm{E}}_{\mathrm{b}}=\Delta m\cdot c^2$

21.0Binding Energy per Nucleon Curve

• The value of binding energy per nucleon decides the stability of the nucleus
• The Greater the binding energy per nucleon ,the more stable the nucleus.
• The maximum for Iron( Fe) is 8.8 Mev, the most stable nucleus.
• For Uranium 7.7Mev it is unstable
• Medium- sized nuclei are more stable than light or heavy nuclei.

22.0Difference between Nuclear Fission and Nuclear Fusion

Required Condition For Nuclear Fusion

• High Temperature
• High Pressure(Density)
• In Nuclear Fission
• Reproduction Factor $(\mathbf{K})=\frac{\text{ Rate of Production of neutrons }}{\text{ Rate of Loss of neutrons }}$
• K = 1, a chain reaction will be steady or sustained.
• K > 1, supercritical this will result in an explosion
• K < 1, Sub critical chain reaction will retard and ultimately stops

23.0Radioactivity and Types of Process

• Radioactivity is a process of spontaneous emission of radiation from the nucleus; only unstable nuclei exhibit this property.

Types of Process

24.0Difference Between alpha, beta and gamma Decay Process

25.0Rutherford's Atomic Model

• Most of the particles went straight through gold foil and produced flashes on the screen.
• Few particles deflected at large angles, and few turned back towards the source itself.
• The entire positive charge and the whole mass of the atom are concentrated in a small centre called the nucleus.
• Electrons cannot deflect the path of particle electrons, which are very light.
• Distance of Closest Approach.

$\left(r_0\right)=\frac{1}{4\pi {ϵ}_0(2ze{)}^2}$ (where, E_K = K.E. of $\alpha$-particle)

Failures of Rutherford's Model

• It could not explain the stability of the atom.
• It could not explain the discrete nature of hydrogen spectra.

26.0Bohr Model of Hydrogen Atom

• An electron in an atom moves around the nucleus in a circular orbit under the influence of Coulomb's force of attraction between electron and nucleus.
• The total energy of the electron revolving in any of the stationary orbits remains constant.
• Electromagnetic radiations are emitted if an electron jumps from a stationary orbit of higher energy E₂ to another stationary orbit of lower energy E₁.

$E_2-E_1=hv$

Drawbacks of Bohr's Model

• This model does not explain the fine structures of spectral lines, Zeeman effect, Stark effect.
• Valid for single electron system.
• Model- based on circular orbit of electrons.
• Not explain the intensity of spectral lines.
• Not explain the doublets obtained in the spectra of some of the atoms.
• This model is semi quantum model it includes two quantum numbers (E and L).

Formulas for Single Electron Species

(a) Radius of Orbits: ${\mathbf{r}}_{\mathbf{n}}=\frac{n^2{h}^2{ϵ}_0}{\pi m{e}^2}$

(b) Velocity of revolving electrons: $\mathbf{V}=\frac{e^2}{2nh{ϵ}_0}$

(c) Frequency of electron in an orbit: $\mathbf{v}=\frac{m{e}^4}{4{ϵ}_0^2{h}^3{n}^3}$

(d) Electron Energy:  $E_n=\frac{-m{e}^4}{8n^2{h}^2{ϵ}_0^2}$

(e) Frequency of Emitted Radiations

$v=\frac{m{e}^4}{8{ϵ}_0^2{h}^3}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

$\frac{1}{\lambda }=\frac{m{e}^4}{8{ϵ}_0^2{h}^{3}c}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$        

$(\therefore \mathbf{R}=\frac{m{e}^4}{8{ϵ}_0^2{h}^{3}c}$, Rydberg Constant = 1.1 x 10⁷m^-1 $)$

$\bar{v}=\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

(f) Electron energy level in Hydrogen atom: $E_n=-\frac{13.6}{n^2}\mathrm{eV}$

27.0Spectral Series of Hydrogen Atom

$\bar{v}=\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Lyman series - This series lies in the ultraviolet region.

(a) First Member: $\frac{1}{\lambda }=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3R}{4}=\lambda =1216\mathring{A}$

(b) Second Member: $\frac{1}{\lambda }=R\left[\frac{1}{1^2}-\frac{1}{3^2}\right]=\frac{8R}{9}=\lambda =1026\mathring{A}$

(c) Limiting Member: $\frac{1}{\lambda }=R\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]=R=\lambda =912\mathring{A}$

Balmer Series- This series lies in the visible region and near the ultraviolet region

(a) First Member: $\frac{1}{\lambda }=R\left[\frac{1}{2^2}-\frac{1}{3_2^2}\right]=\frac{5R}{36}=\lambda =6563\mathring{A}$

(b) Second Member: $\frac{1}{\lambda }=R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3R}{16}=\lambda =4861\mathring{A}$

(c) Third Member: $\frac{1}{\lambda }=R\left[\frac{1}{2^2}-\frac{1}{{\infty }^2}\right]=\lambda =3646\mathring{A}$

Paschen Series-This series lies in the infra- red region.

$\frac{1}{\lambda }=R\left[\frac{1}{3^2}-\frac{1}{n_2^2}\right]$, for the first member the wavelength is 18750 Å

Brackett Series- This series lies in Infra-red region: $\frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{4^2}-\frac{1}{n_2^2}\right]$

Pfund Series- This series lies in Infra-red region: $\frac{1}{\lambda }=R\left[\frac{1}{5^2}-\frac{1}{n_2^2}\right]$

28.0Solved Problems on Modern Physics

Question 1. Power of a radiator is 100 watt and its wavelength( $\lambda$) is 400 nm. Calculate No. of photons emitted in 10 hrs.

Solution:

$\mathrm{n}=\left(5\times 10^{24}{\mathrm{J}}^{-1}{\mathrm{m}}^{-1}\right)P\times \lambda$ 

n = 5 × 10²⁴ × 100 × 4 × 10^–7 = 2 × 10²⁰

N = n × t = 2 × 10²⁰ × 10 × 3600 = 7.2 × 10²⁴

Question 2. A source S₁ is producing 10¹⁵ photon/s of wavelength 5000 Å. Another source S₂ is producing 1.02 × 10¹⁵ photon/s of wavelength 5100 Å. Calculate ratio of power of S₂ to power of S₁.

Solution:

$\mathrm{n}=\frac{P\lambda }{hc}\text{ or }\mathrm{P}=\frac{nhc}{\lambda }$

$\frac{P_2}{P_1}=\frac{n_2}{{\lambda }_2}\times \frac{{\lambda }_1}{n_1}=\frac{1.02\times 10^{15}\times 5000}{5100\times 10^{15}}=\frac{1}{1}$

Question 3. A beam of electromagnetic wave of power 10 watt is incident normally on a surface, which absorbs 40% power and remaining power is reflected. Calculate force exerted on a surface.

Solution:

$\mathrm{F}=\frac{P_i+P_r}{c}=\frac{10+6}{3\times 10^8}=5.33\times 10^{-8}\mathrm{N}$

Question 4. Find the ratio of de-Broglie wavelengths for a proton and an alpha particle, if both have the same speed.

Solution:

$\lambda =\frac{h}{mv}\Rightarrow \lambda \propto \frac{1}{m}$

$\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{m_{\alpha }}{m_p}=\frac{4m_p}{m_p}=\frac{4}{1}$

Question 5. The work function of a metal is 2.3 eV and the wavelength of the incident photon is 4.8 × 10^–7m.  Find maximum kinetic energy of photoelectrons.

Solution:

$\varphi =2.3\mathrm{eV}$, $\lambda$ = 4.8 x 10${}^{-7}$ = 4800 $\mathring{A}$K . $E_{\max }=\frac{hc}{\lambda }-\varphi =\frac{\left(6.62\times 10^{-34}\right)\left(3\times 10^8\right)}{\lambda }$

$K\cdot E_{\max }=\frac{12400}{4800}-2.3=0.28\mathrm{eV}$