Dual Nature of Matter and Radiation

The Dual Nature of Matter and Radiation is a key concept in modern physics, revealing that both matter and light exhibit wave-particle duality. This means particles like electrons and photons can behave as both waves and particles, depending on the experiment. The photoelectric effect, explained by Einstein, demonstrated light's particle nature through photons. Similarly, Louis de Broglie showed that matter, like electrons, also exhibits wave-like properties, described by the de Broglie relation. This duality is central to quantum mechanics and plays a crucial role in understanding physics.

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

3.0Work Function ($\varphi$)

It is the minimum energy an electron requires to escape from a metal surface. It is measured in electron volts(eV).

4.0Einstein’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.

Energy of Photon

• Energy is radiated from the source in the form of small packets called photons. 
• According to Planck's Theory, $\text{Energy}(E)\propto \text{(Frequency)}$

$E=h\nu =\frac{hc}{\lambda }=\frac{12400}{\lambda }eV-A^o$

Linear Momentum of Photon: $p=\frac{E}{c}=\frac{h\nu }{c}=\frac{h}{\lambda }$

Effective Mass of Photon: $m=\frac{E}{c^2}=\frac{hc}{c^2\lambda }=\frac{h}{c\lambda }$

The rest mass of the Photon is always zero.

$R>V{\lambda }_R>{\lambda }_V$

5.0Intensity of Light

$I=\frac{E}{At}=\frac{P}{A}$

S.I Unit: $\frac{Joule}{m^2-s}$ or $\frac{watt}{m^2}$

$I=\frac{N(h\nu )}{At}=\frac{n(h\nu )}{A}$$\left(\therefore n=\frac{N}{t},\text{no. of photons per sec.}\right)$

6.0Radiation Force and Pressure

1. When radiations are falling normally on a perfectly reflecting surface-

Radiation Force $F=\frac{2h}{\lambda }\times \frac{P\lambda }{hc}=\frac{2P}{c}$

Radiation Pressure $P=\frac{F}{A}=\frac{2P}{cA}=\frac{2I}{c}I=PA\left[I=\frac{P}{A}\right]$

2. When radiations are falling normally on a perfectly absorbing surface-

Radiation Force $F=n\frac{h}{\lambda }$

Radiation Pressure $P=\frac{F}{A}=\frac{P}{Ac}=\frac{I}{c}$

7.0Photoelectric 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.

Lenard’s Experiment of Photoelectric Effect

• When light of frequency $\nu$ and intensity I  falls on the cathode, electrons are emitted from it. 
• The electrons are collected by the anode and a current flows in the circuit. This current is called photoelectric current.
• This experiment is used to study the variation of photoelectric current with different factors like intensity, frequency and the potential difference between the anode and cathode.

Saturation 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.

Stopping 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.

Retarding potential: Negative potential of anode with respect to cathode which is less than stopping potential is called retarding potential.

8.0Graph 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

$e{V}_0=h\nu -{\varphi }_0$

$V_0=\left[\frac{h}{e}\right]\nu -\left[\frac{{\varphi }_0}{e}\right]$

Slope $m=\tan \theta =\frac{h}{e}\text{(same for all metals)}$

9.0Einstein 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,

$h\nu =K.E$ 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.

$h\nu ={K.E}_{\text{Max}}+{\varphi }_0$….Einstein Photoelectric Equation

${K.E}_{\text{Max}}=h\nu -{\varphi }_0$

$e{V}_0=h\nu -h{\nu }_0$$(\therefore v_{0}isthethresholdfrequencyforthat{V}_0=0)$

• K.E. and Stopping Potential $v_0$ depend on frequency.
• Kinetic Energy cannot be negative so that $h\nu >h{\nu }_0$
• $\nu >{\nu }_0$ (It means if frequency is less than ‘0’${\nu }_0$ , electrons do not come out.

The Graph between K.E. and Frequency

$(K{)}_{\text{max}}=h\nu -{\varphi }_0$

$slope=m=\tan \theta =h\text{(same for all metals)}$

$({\varphi }_0{)}_B>({\varphi }_0{)}_A$

10.0Quantum Efficiency

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

$Ifquantumefficiencyisx\%\text{ then }{n}_e=\frac{x}{100}\left[\therefore n_{ph}=(5\times 10^{24}{\text{J}}^{-1}\cdot {\text{m}}^{-1})PA\right]$

11.0Matter 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

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

$h=6.63✕10^{3}4J-s$

De-Broglie wavelength for a charged particle

Consider a charged particle of mass 'm' and charge 'q' accelerated from rest through a potential V, then its

1. Kinetic energy K=qV
2. Momentum $p=\sqrt{2mK}=\sqrt{2mqV}$
3. Velocity $v=\sqrt{\frac{2qV}{m}}\left[K=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{\frac{2K}{m}}\right]$
4. de-Broglie Wavelength $\lambda =\frac{h}{\sqrt{2mqV}}$

De-Broglie Wavelength associated with Uncharged Particles

1. Neutron   $\left({\begin{array}{c}n\\ 0\end{array}}^1\right)$

${\lambda }_n=\frac{h}{P_n}=\frac{h}{m_n{v}_n}=\frac{h}{\sqrt{2m_{n}K}}$Here K must be in Joule(J)

${\lambda }_n=\frac{0.2863}{\sqrt{K}}\text{A˚}/\sqrt{\text{eV}}$ Here K must be in eV

Special Note: For thermal neutrons, average kinetic energy is given by $\frac{3}{2}kT$

${\lambda }_n=\frac{h}{\sqrt{2m_n\left(\frac{3}{2}kT\right)}}=\frac{h}{\sqrt{3m_{n}kT}}\approx \frac{25.6}{\sqrt{T}}\text{A˚}/\sqrt{K}$(Temperature is in Kelvin)

2. Gas Molecule: Total kinetic energy of any gas molecule is $\frac{f}{2}kT$ where f=degree of Freedom

But total translatory kinetic energy of any gas molecule is $\frac{3}{2}kT$ because de-Broglie wavelength is associated with translational motion only. Therefore, for any gas molecule.

$\lambda =\frac{h}{\sqrt{2m(\frac{3}{2}kT)}}$ $=\frac{h}{\sqrt{3mkT}}$ (Here T is in Kelvin Scale)

12.0Explanation of Bohr Quantization 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.

$2\pi r=n\lambda$      

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

$mvr=\frac{nh}{2\pi }$This is Bohr quantization Condition

$r_n=(0.529A^o)\frac{n^2}{Z}$ (∴Z=Atomic number)

13.0Sample Questions on Dual Nature of Matter And Radiation

Q-1.A source $S_1$ is producing $10^{15}$ photon/sec of wavelength 5000 Å. Another source $S_2$ is producing $1.02\times 10^{15}$ photon/sec of wavelength 5100 Å. Calculate ratio of power of $S_2$ power of

$S_1.$

Solution:

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

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

Q-2.The energy flux of sunlight reaching on the Earth's surface is 1.388 × 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average length of 550 nm.

Solution:

$\text{Intensity }I=\frac{\text{Power Emitted}}{\text{Area}}=\frac{P}{A}=1.388\times 10^3{\text{W/m}}^2$

$\text{Energy of a Photon, }E=\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3\times 10^8}{550\times 10^{-9}}=3.616\times 10^{-19}\text{J}$

Let n be the total number of photon/area then,

$n=\frac{P/A}{E}=\frac{I}{E}=\frac{1.388\times 10^3}{3.616\times 10^{-19}}=3.83\times 10^{21}{\text{photons/m}}^2\text{s}$

Q-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:

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

Q-4.Find the ratio of de-Broglie wavelengths for a proton and a deuteron, if both have the same kinetic energy.

Solution:

$\lambda =\frac{h}{\sqrt{2mK}}\Rightarrow \lambda \approx \frac{1}{\sqrt{m}}$

$\frac{{\lambda }_p}{{\lambda }_d}=\sqrt{\frac{m_d}{m_p}}\approx \sqrt{\frac{2m_p}{m_p}}=\frac{\sqrt{2}}{1}$

Q-5.Compute the typical de-Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal, which is given to be about 2 ✕ 10-10m .

Solution:

$\lambda =\frac{h}{\sqrt{3mkT}}=\frac{6.6\times 10^{-34}}{\sqrt{3\left(9.1\times 10^{-31}\right)\left(1.38\times 10^{-23}\right)(273+27)}}=6.2\times 10^{-9}\text{m}$

As mean separation between two electrons =$\frac{r}{\lambda }=\frac{2\times 10^{-10}}{6.2\times 10^{-9}}\approx 0.03$