Class 12 Physics Dual Nature of Radiation and Matter Formulas

• Mass of an Electron

$m_e=9.11\times {10}^{-31}\text{kg}$

• Charge of an Electron

$e=1.602\times {10}^{-19}\text{C}$

• Charge to mass (e/m) Ratio of an Electron

$\frac{e}{m_e}=1.76\times {10}^{11}\text{C/kg}$

• Methods of Emission:
  • Thermionic Emission: Free electrons come out of the metal surface due to an excess heating of the metal means more thermal energy supply.
  • Photoelectric Emission: Electrons are emitted from a metal surface due to illumination light on the metal surface.
  • Field Emission: Electron is forced to come out of the metal surface due to a strong electric field.
• Work function (Φ_o): The minimum energy required for the electron emission from the metal surface.

$${\varphi }_0=h{\nu }_0=\frac{hc}{{\lambda }_0}$$

• Threshold frequency: It is minimum cut-off frequency of incident radiation for which photoelectric emission is possible.

$${\nu }_0=\frac{{\varphi }_0}{h}$$

• Intensity of Light: The number of photoelectrons emitted per second is proportional to the intensity of radiation.

$$I=\frac{\text{Power}}{\text{Area}}\text{or}I\propto \text{Number of photons per second}$$

• Saturation Current: It is maximum value of the photoelectric current.

$$I_{\text{sat}}\propto \text{Number of photoelectrons emitted per second}\propto \text{Intensity of incident light}$$

• Stopping potential: The minimum negative potential applied to metal plate for which photocurrent becomes zero.

$$V_0=\frac{h\nu -{\varphi }_0}{e}$$

• Photoelectric equation: It tells about the relation between the work function and kinetic energy

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

• Maximum Kinetic Energy of the photon that came out of the metal surface.

$$K_{\text{max}}=\frac{1}{2}m{v}^2$$

• Momentum of the electron or wave:

$$p\; =\frac{h}{\lambda }=\frac{h\nu }{c}$$

• Energy of photon, which comes out due to radiation:  

$$E=h\nu =\frac{hc}{\lambda }$$

•  Number of photons emitted by the source per second: 

$$N = \frac{P}{E} = \frac{P}{h\nu}$$

• De Broglie wavelength of a photon:

$$\lambda =\frac{h}{p}=\frac{h}{mv}$$