Part C What is the kinetic energy of the emitted electrons when cesium is exposed to...

Question

Question

Part C

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.4×1015Hz?

Express your answer in joules to three significant figures.

Answer 1

Answer #1

We have,

KE = E – E_o = hv –hv_o

Where, h = 6.626 x 10^-34 Js

Given frequency = 1.4 x 10¹⁵ Hz

v_o (Threshold frequency) for cesium = 9.39 x 10¹⁴ Hz

Applying in equation 1, we have

KE = 6.626 x 10³⁴(1.4 x 10¹⁵- 9.39 x 10¹⁴) = 3.054 x 10^-19 J

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

Answer 2

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