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Photoelectric emission is observed from a surface for frequencies `v_(1) and v_(2)` of the incident radiation `(v_(1) gtt v_(2))`. If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio `1 : k`, what will be the theshold frequency `(v_(0))` in term of `v_(1), v_(2)` and k ? |
Answer» In 1st case, K.E. `= h v_(1) - h v_(0) = h (v_(1) - v_(0))` In 2nd case, K.E. `= h v_(2) - h v_(0) = h (v_(2) - v_(0))` Given `(h (v_(1) - v_(0)))/(h(v_(2) - v_(0))) = (1)/(k) or v_(2) -v_(0) = k (v_(1) - v_(0)) = k v_(1) - kv_(0)` or `k v_(0) - v_(0) = kv_(1) - v_(2) or v_(0) (k -1) = kv_(1) = v_(2) or v_(0) = (kv_(1) - v_(2))/(k -1)` |
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