Question

Light of wavelength 500 nm is incident on a metal with work function \( 2.28 \, \text{eV} \). The de Broglie wavelength of the emitted electron is:

• A. \(<2.8 \times 10^{-9} \, \text{m} \)
• B. \(>2.8 \times 10^{-10} \, \text{m} \)
• C. \(<5.2 \times 10^{-12} \, \text{m} \)
• D. \(<2.8 \times 10^{-10} \, \text{m} \)

Hint:

The de Broglie wavelength of an emitted electron can be calculated using the relation \( \lambda = \frac{h}{p} \), where \( p \) is the momentum of the electron.

Answer 1

The Correct Option is A

Step 1: Use the photoelectric equation.
Using the photoelectric equation, the kinetic energy of the emitted electron is given by: \[ K.E. = h \nu - W \] where \( h \nu \) is the energy of the incident photon, and \( W \) is the work function. 

Step 2: Use the de Broglie wavelength formula. 
The de Broglie wavelength \( \lambda \) of the emitted electron is related to its momentum by: \[ \lambda = \frac{h}{p} \] where \( p = \sqrt{2m K.E.} \). 

Step 3: Calculate the de Broglie wavelength. 
The de Broglie wavelength can be found using the above relations, which yields: \[ \lambda<2.8 \times 10^{-9} \, \text{m} \] 
Final Answer: \[ \boxed{< 2.8 \times 10^{-9} \, \text{m}} \]