Question

Light of wavelength 5000 falls on a sensitive plate with photoelectric work function 1.9 eV. The kinetic energy of the photoelectron is:

• A. 0.58 eV
• B. 2.48 eV
• C. 1.24 eV
• D. 1.16 eV

Hint:

$E(eV) \approx 12400 / \lambda$() is a useful shortcut.

Answer 1

The Correct Option is A

Step 1: Concept
Einstein's photoelectric equation: $K.E. = E - \Phi$.
Step 2: Analysis
Energy of photon $E = 12400 / 5000 = 2.48 eV$.
$K.E. = 2.48 - 1.9$.
Step 3: Conclusion
$K.E. = 0.58 eV$.
Final Answer: (A)