Chapter 11: Dual Nature of Radiation and Matter - Comprehensive NEET Physics Notes

1. Photoelectric Effect

1.1 Key Formulae

1. Einstein's Photoelectric Equation:
   $K_{\text{max}}=h\nu -\varphi$
2. • Explanation:
   • • $K_{\text{max}}$: Maximum kinetic energy of the emitted photoelectron.
     • $h$: Planck’s constant ($6.63\times 10^{-34}\text{J s}$).
     • $\nu$ Frequency of the incident light.
     • ${\varphi }_0$​: Work function of the metal, the minimum energy required to emit an electron from the metal surface.
   • Assumption: This equation assumes that the incident light frequency $\nu$ is greater than the threshold frequency ${\nu }_0$ of the metal.
2. Threshold Frequency:
   ${\nu }_0=\frac{{\varphi }_0}{h}$
3. • Explanation:
   • • ${\nu }_0$​: Threshold frequency, the minimum frequency needed for photoemission.
     • ${\varphi }_0$: Work function of the metal.
   • Condition: Photoelectric emission occurs only if $\nu >{\nu }_0$.
3. Stopping Potential:
   $e{V}_0=K_{\text{max}}$
4. • Explanation:
   • • $e$: Charge of the electron ($1.6\times 10^{-19}\text{C}$).
     • $V_0$​: Stopping potential, the potential at which the photoelectric current becomes zero.
   • Application: Used to measure the maximum kinetic energy of photoelectrons.

1.2 Derivations

• Derivation of Einstein's Photoelectric Equation:
• • Start with the concept that energy is quantized in photons.
  • Each photon has energy $h\nu$.
  • When a photon hits the metal surface, an electron absorbs this energy.
  • If the energy exceeds the work function ${\varphi }_0$, the electron is ejected with kinetic energy $K_{\text{max}}=h\nu -{\varphi }_0$​.

1.3 Example Applications

1. Problem:
2. • Find the maximum kinetic energy of electrons emitted from a metal with a work function ${\varphi }_0=2.14\text{eV}$ when light of frequency $\nu =6\times 10^{14}\text{Hz}$is incident.
   • Solution:
   • • $K_{\text{max}}=h\nu -{\varphi }_0=(6.63\times 10^{-34}\text{J s})\times (6\times 10^{14}\text{Hz})-2.14\text{eV}$
     • $K_{\text{max}}=3.98\times 10^{-19}\text{J}-2.14\times 1.6\times 10^{-19}\text{J}$
     • $K_{\text{max}}=0.57\times 10^{-19}\text{J}$
2. Problem:
3. • Calculate the stopping potential if the maximum kinetic energy of photoelectrons is $2\times 10^{-19}\text{J}$.
   • Solution:
   • • $V_0=\frac{K_{\text{max}}}{e}=\frac{2\times 10^{-19}\text{J}}{1.6\times 10^{-19}\text{C}}=1.25\text{V}$

1.4 Common Mistakes

• Mistake: Confusing frequency with wavelength when applying the photoelectric equation.
• • Tip: Remember, frequency $\nu$ is directly used in the equation $K_{\text{max}}=h\nu -{\varphi }_0$. Wavelength $\lambda$ should be converted using $\nu =\frac{c}{\lambda }$​.
• Mistake: Assuming that increasing the intensity of light will increase the kinetic energy of emitted electrons.
• • Tip: Kinetic energy depends on frequency, not intensity.

Quick Recap

• Einstein's Photoelectric Equation: $K_{\text{max}}=h\nu -{\varphi }_0$
• Threshold Frequency: ${\nu }_0=\frac{{\varphi }_0}{h}$
• Stopping Potential: $e{V}_0=K_{\text{max}}$
• Important: Photoemission occurs only if $\nu >{\nu }_0$​.

NEET Problem-Solving Strategy

• Tip: Always start by checking if the frequency of incident light is above the threshold frequency before applying Einstein’s equation.
• Time Management: For questions involving the photoelectric effect, quickly identify the relevant formula based on what is given: kinetic energy, work function, or stopping potential.

Practice Questions

1. Question: Light of frequency $5\times 10^{14}\text{Hz}$ is incident on a metal with work function $2.5\text{eV}$. What is the stopping potential?
2. • Solution: Apply $e{V}_{}$