Question

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.7×10^{15} Hz?

Answer 1

jog

Answer 2

0.7889

Answer 3

kglig

Answer 4

[AI-generated answer — Explain Bot]

To calculate the kinetic energy of the emitted electrons, we can use Einstein's photoelectric equation:

E = hf - φ

Where:
E is the kinetic energy of the emitted electrons
h is Planck's constant (6.626 × 10^-34 J s)
f is the frequency of the UV rays (1.7×10^15 Hz)
φ is the work function of the material (in this case, cesium)

First, let's find the value of the work function for cesium. The work function is the minimum energy required to remove an electron from an atom or material. It can be experimentally determined or given in the problem.

Once we have the work function value, we can substitute the given values into the equation to find the kinetic energy.

Do you have the value of the work function for cesium?