COMEDK Exam 2019 Solved Paper

Threshold wavelength of the metal, λ0=5200Å
=5200×10−10m
Frequency of the radiation, v0=‌

c

λ

=‌‌

3×108

5200×10−10

=‌0.57×1015‌Hz
=‌5.7×1014‌Hz
The frequency of infrared radiation is lower than the threshold frequency of the metal. Hence, the surface of the metal is illuminated with monochromatic radiation which is greater than 5.7×1014‌Hz.
The 50W UV-lamp is suitable for this purpose.