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Photoelectric emission is observed from a metal surface with incident frequencies `v_(1) and v_(2)` where `v_(1) gt v_(2)`. If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio `2 : 1`, then the threshold frequency `v_(0)` of the metal is
A. `v_(1) = v_(2)`
B. `(v_(1) - v_(2))/(h)`
C. `2 v_(1) - v_(2)`
D. `2v_(2) - v_(1)`

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Correct Answer - D
`hv_(1) = hv_(0) + E` (K.E)..(i)
`hv_(2) = hv_(0) + (1)/(2) E or 2 hv_(2) = 2 hv_(0) + E`...(ii)
Eqn. (i) - Eqn (ii) gives `hv_(1) - 2hv_(2) = hv_(0) - 2 hv_(0)`
or `v_(1) - 2 v_(2) = - v_(0) or v_(0) = 2 v_(2) - v_(1)`

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