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

User Joe Thor
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1 Answer

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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.
User Rick Baker
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