1. Photons are the packets of energy which are emitted by a source of radiation and travel in straight line in vacuum with the speed of light.
  2. Energy of photon (E)=hf=\dfrac{hc}{\lambda}

  3. Momentum of photon (P)=\dfrac{h}{\lambda}

  4. When a photon collides with material, total energy and total momentum of photon remains constant.

  5. Number of photons emitted per second or falling per second is given by

    \dfrac{n}{t} = \dfrac{\text{Power} (P)}{\text{Energy of each photon} (E)} = \dfrac{P}{\dfrac{hc}{\lambda}} = \dfrac{P \lambda}{hc}

  6. When photon travels through different mediums, its velocity changes because of change in wavelength whereas its frequency remains constant.

  7. Rest mass of photon is zero i.e. energy of photon is totally kinetic.

  8. Dynamic mass of photon is m=\dfrac{h}{\lambda c} (since, E=mc^2 and E=\dfrac{hc}{\lambda}

Photoelectric effect

  1. \to When electromagnetic wave/radiation of suitable energy falls on a metallic surface, electrons are emitted from a metal surface. This effect is photoelectric effect and emitted electrons are called photoelectrons.

    \to Albert Einstein verified photoelectric effect in 1901 A.D.

    \to Photoelectric effect verifies particle nature of light.

    \to Photoelectric effect is based on principle of conservation of energy.

    \to In photoelectric effect, light energy is converted to electrical energy.

  2. \textbf{Work Function} (\phi)

    \to The minimum energy of incident radiation required to just emit electrons from metallic surface is work function i.e. \phi=hf_0=\dfrac{hc}{\lambda_0} (f_0 and \lambda_0 are called threshold frequency and threshold wavelength respectively.)

    \to Photoelectric effect occurs if f>f_0 and \lambda< \lambda_0 , where f and \lambda are frequency and wavelength of photons. Einstein’s photoelectric equation

    \to If a photon of frequency ‘f’ is incident on a photosensitive metal surface, the energy of photon is spent in two ways:

    i. A part of the energy of photon is used in liberating electrons from metal surface against work function (\phi) .

    ii. The rest of energy of photon is used in liberating the kinetic energy to the emitted photoelectrons i.e.

    E_{photons} = \psi + KE_{\text{emitted electrons}}

    hf = hf_o + \dfrac{1}{2}mv_{max}^2

    \dfrac{hc}{\lambda} = \dfrac{hc}{\lambda _o} + \dfrac{1}{2}mv_{max}^2

  3. \textbf{Stopping potential} (V_o)

    \to The minimum value of negative potential that should be applied against emitted electron to make photoelectric current zero is stopping potential. Mathematically, eV_0=\dfrac{1}{2}mv_{max}^2 =>V_0 \propto v_{max}^2

    \to It is different for different metals.

    \to It is independent upon intensity of incident light but depends on the frequency of incident radiation i.e. V_0f

De-Broglie wave (Matter Wave)

  1. A moving particle exhibits both wave and particle nature.

  2. Electromagnetic wave shows both wave nature and particle nature.

  3. Louis-Victor De-Broglie explained wave-particle duality in 1924 A.D.

  4. De-Broglie wave in some cases:

    CasesMoving particle with 'v' velocityMoving particle with 'KE' kinetic energyaccelerated charge with 'V' potentialGas molecules at 'T' temperature
    De-Broglie wavelength\lambda = \dfrac{h}{p} = \dfrac{h}{mv}\lambda = \dfrac{h}{\sqrt{2m \times KE}}\lambda = \dfrac{h}{\sqrt{2m \times qV}}\lambda = \dfrac{h}{\sqrt{3m \times kT}}

Heisenberg’s Uncertainty Principle

  1. t was given by Werner Heisenberg in 1927 A.D.

  2. It states “It is impossible to measure both position and momentum of particles simultaneously.”

  3. If \Delta P and \Delta X denotes uncertainty in simultaneous measurements of momentum and position of particles respectively then

    \Delta X \times \Delta P \geq \dfrac{h}{4\pi}

    It also holds for simultaneous measurement of energy and time

    (\Delta E \times \Delta t \geq \dfrac{n}{4\pi}) and simultaneous measurement of angular momentum and angular displacement (\Delta L \times \Delta 0 \geq \dfrac{h}{4\pi})

Some shortcut tricks to solve the problems

  1. When two electrons emitting photons of energy ‘E_1’ and ‘E_2’ gets incident on a metal of work function (𝜙), then ratio of

    I. Kinetic energy of emitted electrons \dfrac{KE_1}{KE_2}=\dfrac {E_1-\phi}{E_2-\phi}

    II. Velocity of emitted electrons \dfrac{V_1}{V_2}=\sqrt{\dfrac{E_1-\phi}{E_2-\phi}}