Nuclear Physics and RadioactivityHard
Question
In a photoelectric experiment, the collector plate is at 2.0V with respected to the emitter plate made of copper ( = 4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200nm.
Options
A.The minimum kinetic energy of the photoelectrons reaching the collector is 0.
B.The maximum kinetic energy of the photoelectrons reaching the collector is 3.7 eV
C.If the polarity of the battery is reversed than answer to part A will be 0
D.If the polarity of the battery is reversed than answer to part B will be 1.7 eV
Solution
Kmax = 
Kmax = 1.7eV at anode
Kmax = (1.7 + 2) eV at anode
If polarity is revered, no e- reach at collector
Kmax = 1.7eV at anode
Kmax = (1.7 + 2) eV at anode
If polarity is revered, no e- reach at collector
Create a free account to view solution
View Solution FreeMore Nuclear Physics and Radioactivity Questions
The figure indicates energy level diagram of an atom and the origin of five spectral lines in emission spectra. Which of...A nucleus with Z = 92 emits the following in a sequence; α, α, β-, β-, α, α, α, α...Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot...Which of the following is/are evidence(s) for the existence of neutrons in an atom ?...Who discovered the nuclear fission :-...