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What is the kinetic energy of the emitted electrons when cesium isexposed to UV rays of frequency 1.4x10^15Hz?

Light energy
() Electron emitted? Electron
()
3.87 no

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Lelia Lubowitz
Lelia LubowitzLv2
28 Sep 2019
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Related questions

what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency of 1.2 x 10^15 hz? Express your answer in joules to three significant figures.

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.3×1015Hz?

By using the following formula, you can calculate the kinetic energy of the emitted electron:

KE=E−E0=hν−hν0

where h=6.63×10−34 J⋠s is Planck's constant, ν=1.3×1015Hz is the given frequency, and ν0=9.39×1014 Hz is your answer from Part A.

I am getting ridiculous numbers and can't seem to get it right.


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