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Lady_Fox [76]
3 years ago
8

what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.4×1015hz?

Chemistry
1 answer:
Anika [276]3 years ago
7 0
<span>The equation you used is KE=hv-hv0, where h=6.63*10^-34 (constant). You multiply h by 1.5*10^15. Multiply h by the threshold freq of cesium (from part A). Subtract the second answer from the first answer, and you get the kinetic energy. Hope this helps.</span>
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