What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.80*1015Hz?

The kinetic energy of emitted electrons when cesium is exposed to UV rays of frequency 1.80 * 10 ^15 is 3.054 x 10 ^ - 19 J.

Explanation:

To calculate the kinetic energy of emitted electrons,

It is given that the frequency is 1.80 * 10^15 Hz

We have,

KE = E - Eo = hv - hvo

Where, h = 6.626 x 10^ - 34 Js

Given frequency = 1.4 x 10 ^ 15 Hz

vo (Threshold frequency) for cesium = 9.39 x 10 ^ 14 Hz

Applying in equation,

we get

KE = 6.626 x 10^34 (1.4 x 10^15 - 9.39 x 10^14)

KE = 3.054 x 10^-19 J

[Note: Here, threshold frequency of Cesium is not provided. Apply the correct threshold frequency from the part A]