Q. The photoelectric threshold wavelength of silver is $3250 \times 10^{-10}$ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10}$ m is : (Given h = $4.14 \times 10^{-15}$ eVs and c = $3 \times 10^8$ $ms^{-1}$)
A) ≈ $6 \times 10^5$ $ms^{-1}$
B) ≈ $0.6 \times 10^6$ $ms^{-1}$
C) ≈ $61 \times 10^3$ $ms^{-1}$
D) ≈ $0.3 \times 10^6$ $ms^{-1}$
Explanation
[Solution] $KE = hc(1/\lambda - 1/\lambda_0) = 4.89 - 3.82 = 1.07$ eV, so $v = \sqrt{2KE/m} \approx 6 \times 10^5$ m/s. Options (1) and (2) are the same value, so both were accepted. Answer (1) / (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.