The threshold frequency for a metal is 8.0×10 to the power 14 per sec what is the k. E. Of an electr mp on emitted having frequency = 1× 10 to the power 15 per sec.
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Answer:
0.82 eV
Explanation:
KE= hv-hvO
= h(v-v0)
6.636×10^-34(1×10^15- 0.8×10^15)
=6.636×10^-34 × 0.2×10^15
=1.32×10^-19 OR
1.32×10^-19/1.6×10^-19
=0.82eV
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