Question

Explain de Broglie wavelength. Obtain an expression for de Broglie wavelength of wave associated with material particles. The photoelectric work function for a metal is 4.2 eV. Find the threshold wavelength.

Hint:

de Broglie hypothesis unifies wave-particle duality; threshold wavelength \( \lambda_0 = \frac{h c}{\phi} \).

Answer 1

Explanation: de Broglie wavelength (\( \lambda \)) is the wavelength associated with a material particle, proposing that matter has wave-like properties. It is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck’s constant and \( p \) is momentum.
Derivation: For photons, \( E = h \nu = pc \), where \( \lambda = \frac{c}{\nu} \Rightarrow p = \frac{h}{\lambda} \Rightarrow \lambda = \frac{h}{p} \).
de Broglie extended this to particles: \( \lambda = \frac{h}{p} = \frac{h}{m v} \), where \( m \) is mass, \( v \) is velocity.
Threshold Wavelength: Threshold frequency \( \nu_0 = \frac{\phi}{h} \), where \( \phi = 4.2 \, \text{eV} = 4.2 \times 1.6 \times 10^{-19} \, \text{J} = 6.72 \times 10^{-19} \, \text{J} \).
\( h = 6.626 \times 10^{-34} \, \text{J s} \).
\[ \nu_0 = \frac{6.72 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 1.014 \times 10^{15} \, \text{Hz}. \] \[ \lambda_0 = \frac{c}{\nu_0} = \frac{3 \times 10^8}{1.014 \times 10^{15}} \approx 2.96 \times 10^{-7} \, \text{m} = 2960 \, \text{Å}. \] Answer: de Broglie wavelength \( \lambda = \frac{h}{p} \); threshold wavelength = 2960 Å.