Answered: 1. A photoelectric experiment with… | bartleby

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A photoelectric experiment with cesium yields stopping potentials of 0.95 V and
546.1 nm, respectively. Using only these numbers
1.
0.38 V for X
together with the values of the speed of light (3 × 10$ m/s) and the electron charge (e =
-1.6 x 10-19 C), find
435.8 nm and A
(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.

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