Question

In a photoelectric experiment, light of wavelength 800 nm produces photoelectrons with de Broglie wavelength 1 nm, and 400 nm light gives 0.5 nm. Find work function of the metal.

• A. 1.03 eV
• B. 0.53 eV
• C. 2.03 eV
• D. 4.03 eV

Hint:

Use energy conservation: \( E = \phi + K.E. = hc/\lambda \)

Answer 1

The Correct Option is A

From Einstein’s equation: \[ hc/\lambda = \phi + p^2/2m \Rightarrow \phi = hc/\lambda - p^2/2m \] Using both cases and converting units, after simplifying, we find: \[ \phi \approx 1.03\ \text{eV} \]