A photoelectric experiment with cesium yields stopping potentials of 0.95 V and 0.38 V for A = 435.8 nm and ) = 546.1 nm, respectively. Using only these numbers together with the values of the speed of light (3 × 10° m/s) and the electron charge (e = -1.6 × 10-19 C), find (a) (b) (c) the work function in eV for cesium, the value for Planck's constant h, and the cutoff frequency, below which no electrons would be emitted.
A photoelectric experiment with cesium yields stopping potentials of 0.95 V and 0.38 V for A = 435.8 nm and ) = 546.1 nm, respectively. Using only these numbers together with the values of the speed of light (3 × 10° m/s) and the electron charge (e = -1.6 × 10-19 C), find (a) (b) (c) the work function in eV for cesium, the value for Planck's constant h, and the cutoff frequency, below which no electrons would be emitted.
Related questions
Question
How to solve this question
Transcribed Image Text:A photoelectric experiment with cesium yields stopping potentials of 0.95 V and
0.38 V for A = 435.8 nm and A = 546.1 nm, respectively. Using only these numbers
together with the values of the speed of light (3 × 10° m/s) and the electron charge (e =
-1.6 × 10-19 C), find
(a)
the work function in eV for cesium,
(b)
the value for Planck's constant h, and
(c)
the cutoff frequency, below which no electrons would be emitted.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 5 steps with 5 images
