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Calculate the threshold freqency of the metalfrom the photoelectronsare emitted with zero velocity when exposed to radiation of wavelength 6800 `Å` Hint `: KE = (1)/(2) mv^(2)` `hv= hv_(0) + KE, KE=0` `v= (c ) /( lambda)` |
Answer» `4.41 xx 10^(14) s^(-1)` | |